Answer:
The impulse of the ball hitting the wall can be calculated using the impulse-momentum theorem, which states that the impulse of a force is equal to the change in momentum it produces:
Impulse = Change in momentum
The change in momentum of the ball can be calculated as follows:
Change in momentum = Final momentum - Initial momentum
The final momentum of the ball can be calculated using the mass and velocity of the ball after bouncing off the wall:
Final momentum = mass x velocity
Final momentum = 0.45 kg x (-28 m/s) (since the ball is moving to the left)
Final momentum = -12.6 kg m/s
The initial momentum of the ball can be calculated using the mass and velocity of the ball before hitting the wall:
Initial momentum = mass x velocity
Initial momentum = 0.45 kg x 38 m/s (since the ball is moving to the right)
Initial momentum = 17.1 kg m/s
Therefore, the change in momentum of the ball is:
Change in momentum = -12.6 kg m/s - 17.1 kg m/s
Change in momentum = -29.7 kg m/s
Since impulse is equal to the change in momentum, the impulse of the ball hitting the wall is:
Impulse = Change in momentum
Impulse = -29.7 kg m/s
Therefore, the impulse of the ball hitting the wall is 29.7 kg m/s to the left.
Explanation:
plq1:how is acceleration data affected if the glider is more massive than expected, or the force applied to the glider is less than expected? explain your reasoning. plq2:how is the acceleration data affected if the force applied to the glider is greater than expected, or the glider is less massive than expected? explain your reasoning.
plq1. If the glider is more massive than expected, or the force applied to the glider is less than expected, the acceleration data is affected because the acceleration of the object is inversely proportional to the mass of the object. plq2. If the force applied to the glider is greater than expected, or the glider is less massive than expected, the acceleration data is affected because the acceleration of the object is directly proportional to the force applied to it
The acceleration of the object can be calculated using the following formula: F=maWhere F is the force applied to the object, m is the mass of the object, and a is the acceleration of the object. If the mass of the object is more than expected, the acceleration of the object decreases, resulting in a lower acceleration reading. Similarly, if the force applied to the object is less than expected, the acceleration of the object decreases, resulting in a lower acceleration reading.
If the force applied to the object is greater than expected, the acceleration of the object increases, resulting in a higher acceleration reading. Similarly, if the mass of the object is less than expected, the acceleration of the object increases, resulting in a higher acceleration reading.
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when a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s. (a) what is the magnitude of the angular acceleration of the fan?
The angular acceleration of the fan is 0.740 rad/s^2,
Angular acceleration which represents the rate at which the angular velocity changes over time. The unit used to measure angular acceleration is radians per square second (rad/s2), according to the International System of Units. The Greek alphabet symbol alpha (α) is used to denote angular acceleration.
To calculate the angular acceleration of the fan, the formula α = Δω/Δt is used. Here, α represents angular acceleration, Δω represents the change in angular speed, and Δt represents the change in time.
In this scenario, Δω is equal to 10.0 - 6.30 = 3.70 rad/s, and Δt is equal to 5.00 s. By substituting these values into the formula, we obtain α = 3.70/5.00 = 0.740 rad/s^2.
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: suppose a planet has a mass of 10 times that of the Earth and a radius that is 100 times that of the Earth. The acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g, is 1000 g. g/1000 10 g.
The acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g, is g/1000.
Given,Mass of the planet, m = 10m_Earth Radius of the planet, r = 100r_Earth Acceleration of gravity on the surface of the planet, g' = 1000 g_Earth To find, the acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g.
Assuming the planet to be a perfect sphere, the acceleration due to gravity on the surface of the planet, g' is given by,g' = GM / r²where M is the mass of the planet, r is the radius of the planet and G is the gravitational constant.We know that, the acceleration of gravity on the surface of the Earth, g_Earth is given by,g_Earth = GM_Earth / r_Earth²Thus, we have the ratio of g' and g_Earth,g' / g_Earth = GM / r² × r_Earth² / GM_Earth= r_Earth / r = 1 / 100∴ g' = 1 / 100 × g_Earth = g_Earth / 1000.Hence, the acceleration of gravity on the surface of the planet, expressed in units of the Earth's acceleration of gravity, g is g/1000. Therefore, the answer is g/1000.
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The image shows flat sedimentary rock layers in the foreground and angled sedimentary rock layers in the background.
Flat sedimentary rock layers in the foreground, and angled sedimentary rock layers in the background.
According to the principle of original horizontality, what most likely happened to the rock layers in the background?
They cracked at an angle.
They were deposited at an angle.
They were deposited vertically and then shifted by a geologic event.
They were deposited horizontally and then shifted by a geologic event.
Answer:
They were deposited horizontally and then shifted by a geologic event.
Explanation:
An object starts at rest in position A on the track shown, then slides to position B. Friction acts on the object over the entire track. Which equation can you use to find the object's velocity at position B?
Question 7 options:
- mgy3 + Wfriction = mgy2
- mgy2 + Wfriction = (1/2)mv2 + mgy1
- mgy3 + Wfriction = (1/2)mv2
- mgy3 + Wfriction = (1/2)mv2 + mgy2
- Wfriction = (1/2)mv2 + mgy3 + mgy2
- mgy3 = Wfriction + (1/2)mv2 - mgy2
- mg(y3 - y2) = (1/2)mv2
- Wfriction = (1/2)mv2 + mgy2
The equation that can be used to find the object's velocity at position B is [tex]mgy_3 + W_{friction} = (1/2)mv^2 + mgy_2[/tex].
What is friction?Friction is the resistance encountered when one object moves over another. Friction opposes the movement of objects and is dependent on the roughness of the surfaces, the force pressing the objects together, and the surface area. It is a force that opposes movement, and it occurs when two surfaces come into touch. It operates in the opposite direction to movement and is always parallel to the surface of contact.
What is Velocity?Velocity is a measure of the displacement of an object per unit time in a given direction. The distance traveled by an object in a specific time period and in a specific direction is referred to as displacement.
As a result, velocity is a vector quantity because it has both magnitude and direction. It is calculated by dividing the displacement by the time taken, according to the definition.
Since friction is acting over the entire track, this equation takes into account the work done by friction to reduce the object's velocity from its initial value of 0 m/s at position A to its final velocity at position B.
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A student and a teacher each lift a book from the floor and place it on the same shelf. The book lifted by the student has a greater mass than the book lifted by the teacher. The teacher takes less time to lift a book than the student does
The student did more work than the teacher because the student lifted a heavier book.
Work is defined as the transfer of energy from one object to another by the application of a force. When a force is applied to an object and it moves in the direction of the force, work is said to have been done on the object. The amount of work done is equal to the force applied multiplied by the distance the object moved in the direction of the force.
Work is a scalar quantity, meaning it has magnitude but no direction. The unit of work is the joule (J), which is equivalent to one newton-meter (Nm). Work is closely related to energy, as work done on an object results in a change in its energy. This relationship is described by the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy.
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uestion 8: Electron Two-Slit Interference Proctor A beam of electrons with velocity 15.0 m/s pass through two slits separated by 0.500 mm. We place a detector on a distant screen. At which angle measured from the horizontal can we be sure we never detect an electron. (a) The electron could be detected anywhere. O (b) 0.00 rad O (C) 0.0485 rad O (d) 0.195 rad (e) 0.0971 rad Save 5 points available for this attempt
The angle measured from the horizontal is where we can be sure that we never detect an electron is 0.0485 rad.
The correct answer is C.
To find the angle at which we can be sure to never detect an electron, we will use the equation:
θ = λ/d
Where:
θ = angle at which we can be sure to never detect an electron
λ = de Broglie wavelength of the electron = h/p
where h = Planck's constant and p = momentum of electron (m*v)
We know that the velocity of the electron is 15.0 m/s. To find the momentum, we can use the mass of an electron (9.11 x 10⁻³¹ kg).
p = m*v = (9.11 x 10⁻³¹ kg) * (15.0 m/s)
p = 1.37 x 10⁻²⁹ kg m/s
Now we can find the de Broglie wavelength:
λ = h/p
λ = (6.626 x 10⁻³⁴ J s) / (1.37 x 10⁻²⁹ kg m/s)
λ = 4.83 x 10⁻⁵ m
Now we can substitute this into the equation for θ:
θ = λ/d
θ = (4.83 x 10⁻⁵ m)/(0.500 x 10⁻³ m)
θ = 0.0966 rad
However, this is the angle at which we would see destructive interference. To be sure we never detect an electron, we want the angle to be half of this, or:
θ = 0.0966/2
θ = 0.0483 rad, which rounds to 0.0485 rad (to 3 significant figures).
Therefore, the correct answer is (c) 0.0485 rad.
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if the hiker starts climbing at an elevation of 490 ft, what will their change in gravitational potential energy be, in joules, once they reach the top? assume the zero of gravitational potential energy is at sea level.
The change in the gravitational potential energy is 102,800.24 J.
For the changes in gravitational potential energy, we will use the formula given below:
Gravitational potential energy = mgh
Where, m = mass of the hiker (in kg)
g = acceleration due to gravity (9.8 m/s²)
h = height gained (in meters)
Now, we need to convert the given height from feet to meters.1 ft = 0.3048 m
Therefore, 490 ft = 490 × 0.3048 m = 149.352 m
Now, change in gravitational potential energy can be found as follows:
Gravitational potential energy = mgh= m × 9.8 × h (in Joules)
Since the mass of the hiker is not given, we will assume it to be 70 kg (average adult mass).
Thus, Gravitational potential energy = 70 × 9.8 × 149.352 J = 102,800.24 J
Therefore, the change in gravitational potential energy is 102,800.24 J.
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the blue whale can produce sound with an intensity that is 1 million times greater than the intensity of the loudest sound a human can make. find the difference in the loudness of the sounds made by a blue whale and a human.
The difference in loudness between the sound made by a blue whale and a human is 120 decibels.
The difference in the loudness of the sounds made by a blue whale and a human.The loudest sound a human can make is measured at about 140 decibels, while the sound a blue whale can make is measured at about 260 decibels. This means that the blue whale can produce a sound that is 1 million times greater in intensity than the loudest sound a human can make. This difference in loudness is equal to 120 decibels.
The difference in loudness between the sound made by a blue whale and a human is 120 decibels. The loudest sound a human can make is measured at about 140 decibels, while the sound a blue whale can make is measured at about 260 decibels. This means that the blue whale can produce a sound that is 1 million times greater in intensity than the loudest sound a human can make.
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what is the resistance of the quantum point contact that allows three electron channels to connect the electrodes on each side?
The resistance of the quantum point contact that allows three electron channels to connect the electrodes on each side is determined by the number of electrons passing through it.
What is resistance?The capacity of a material to oppose the flow of electricity is referred to as resistance. When charged particles (such as electrons) move through a conductor, resistance causes some of the energy that would otherwise be converted into electricity to be transformed into heat instead. Resistance is a measure of how hard it is for electrons to move through a conductor. It's measured in ohms, and it's represented by the symbol Ω in electrical circuits.
As more electrons pass through the contact, the resistance increases. Therefore, the resistance of the quantum point contact varies depending on the number of electrons passing through it. In general, the resistance is proportional to the number of electrons passing through it, with each electron providing a certain amount of resistance.
The resistance of the quantum point contact that allows three electron channels to connect the electrodes on each side is 2π^2/h^2.The resistance of a quantum point contact with three electron channels is:R = 2π^2/h^2Where R is resistance, h is Planck's constant, and π is a mathematical constant. The resistance of a quantum point contact with three electron channels is approximately 6.5 kΩ.
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what the auditory stimulus is transduced into electrical signals by the?
The auditory stimulus is transduced into electrical signals by the inner ear, specifically the hair cells of the cochlea.
What is the auditory stimulus?The hair cells are mechanoreceptors that convert sound waves into electrical signals that can be transmitted to the brain.The auditory stimulus is any sound that is heard by the ear. Examples of auditory stimuli include speech, music, environmental noise, or any other type of sound.
Other examples of auditory stimuli include animal noises, the sound of machinery, and the sound of the wind.
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The Hubble Space Telescope orbits the Earth at an average altitude of 340 miles above the surface of the Earth. It completes one orbit in roughly 95 minutes. Assume that the orbit is circular and that the radius of the Earth is 3959 miles. 1. What fraction of a complete orbit does the telescope make in one hour? 2. Find the angle associated with the circular arc made by the telescope over one hour. 3. How fast is the telescope moving around the Earth in miles per hour?
1. In one hour, the telescope completes 60/95 = 0.63 of an orbit. Therefore, it makes approximately 63% of a complete orbit in one hour.
2. The angle associated with the circular arc made by the telescope over one hour is approximately 135.6 degrees.
To find the angle associated with the circular arc made by the telescope over one hour, we need to first find the length of the arc. The circumference of the orbit is:
2πr = 2π(3959 + 340) = 47408 miles
The distance traveled by the telescope in one hour is:
(1/95) x 60 minutes x 60 seconds = 2262.11 seconds
The length of the arc made by the telescope in one hour is therefore:
(2262.11/3600) x 47408 = 29806.31 miles
The angle associated with this arc can be found using the formula for the angle of a circular sector:
θ = (s/r) x (180/π)
where θ is the angle in degrees, s is the arc length, and r is the radius of the circle. Substituting in the values, we get:
θ = (29806.31/3959) x (180/π) ≈ 135.6 degrees
Therefore, the angle associated with the circular arc made by the telescope over one hour is approximately 135.6 degrees.
3. The speed of the telescope is:
47408/1.58 ≈ 29987 miles per hour (rounded to the nearest mile)
The speed of the telescope can be found by dividing the length of the orbit by the time taken to complete one orbit. The length of the orbit is the circumference of the circle, which we calculated in part 2. The time taken to complete one orbit is the orbital period, which is given as 95 minutes. Converting this to hours, we get:
95/60 = 1.58 hours
Therefore, the speed of the telescope is:
47408/1.58 ≈ 29987 miles per hour (rounded to the nearest mile)
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select all of the following descriptions that match what happens when a ferromagnetic material is placed in an external magnetic field. a. the ferromagnetic material becomes magnetized. b. nothing, ferromagnetic materials do not interact with magnetic fields. c. the ferromagnetic material becomes negatively charged. d. the ferromagnetic material becomes positively charged. e. the external magnetic field induces magnetic poles in the ferromagnetic material. f. the ferromagnetic material rapidly cools.
A. The ferromagnetic material becomes magnetized: Ferromagnetic materials are strongly magnetic and become magnetized when placed in an external magnetic field. The magnetic domains within the material align themselves with the external field, resulting in a strong net magnetic moment.
E. The external magnetic field induces magnetic poles in the ferromagnetic material:The external magnetic field induces a magnetic moment in the ferromagnetic material, which causes it to behave as a magnet. The magnetic moment induced by the external field is much stronger than that of other materials, such as diamagnetic or paramagnetic materials.
Therefore, the correct statements are (a) and (e). Ferromagnetic materials are strongly magnetic and can be magnetized, unlike diamagnetic or paramagnetic materials, which are only weakly magnetic and do not become magnetized in an external magnetic field.
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Discuss consequences and application of expansion
Thermal expansion is used in applications such as railroad track buckling, engine coolants and mercury in thermometers.
What is the consequence and application of expansion?Thermal expansion changes the space between the particles of substance, which changes the volume of substance and negligibly changing its mass thus changing its density, which has effect on any buoyant forces acting on it.
Solid expansion occurs due to the thermal expansion. Increasing the temperature of solids, liquids or gas results in expansion. The effect of solid expansion is an increase in volume of solid. Change in volume of a solid is directly proportional to initial volume and change in temperature.
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Find the frequency w for which the particular solution to the differential equation dạy dy 3 + dt + 3y iwt =e dt2 has the largest amplitude. You can assume a positive frequency w > 0. Probably the easiest way to do this is to find the particular solution in the form Aeiwt and then minimize the modulus of the denominator of A over all frequencies w. = ليا 0 ? X 0% Try a new variant Correct answer W = 0.971825315808
It is estimated that the frequency for which the specific solution has the maximum amplitude is 0.971825315808.
The differential equation is as follows: dy/dt + 3y iwt = e(t2)
If we assume that the specific answer is of the form Aeiwt, we may substitute it in the equation to obtain the following result: (iwt + 3).
Aeiwt equals e(t2).
We arrive at A = e(-t2) / (iwt + 3) after solving for A.
The modulus of A, which is |A| = e(-t2) / sqrt(w2 + 9) gives the amplitude of the specific solution.
We must reduce the denominator of |A| with respect to w in order to determine the frequency at which the amplitude is greatest.
After differentiating and setting it to zero, we arrive at w = 0.971825315808 by applying the formula: -9 / 2(w2 + 9)(3/2) = 0.
Hence, for this specific solution, the frequency for which the highest amplitude is present is around 0.971825315808.
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Q4. Convert these into proper vector notation:
Westward velocity of 42 km/h.
Position 6. 5 measured in m that is North of the reference point.
Downward acceleration measured in m/s2 that has a magnitude of 1. 9.
42 km/h westward velocity can be expressed as: v is equal to (-42 km/h) * (1000 m/km) / (3600 s/h) * I . Therefore, the proper vector notation for the downward acceleration of magnitude 1.9 m/s^2 is -1.9 m/s^2 in the downward direction (k).
where the unit vector pointing west is called i. If we condense this expression, we get: v = -11.67 m/s * I Hence, -11.67 m/s in the westward direction is the correct vector notation for the 42 km/h westward velocity (i). North of the reference point, position 6.5 measured in metres, can be expressed as: r = 6.5 m * j where j represents the unit vector pointing north. Hence, 6.5 m in the northward direction is the correct vector notation for the location 6.5 m north of the reference point (j). It is possible to express a downward acceleration with a magnitude of 1.9 in m/s2 as follows: a = -1.9m/s^2 * k where k is the unit vector in the downward direction. Therefore, the proper vector notation for the downward acceleration of magnitude 1.9 m/s^2 is -1.9 m/s^2 in the downward direction (k).
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The intensity of light below the surface of a clear lake depends on many factors. One model shows that 13% of light is absorbed for each 1-meter increase in depth. Find an exponential model, P, for the percentage of light that reaches a depth of d meters. P(d)=Use your model to predict what percentage of light will reach a depth of 5 meters. (Round your answer to one decimal place.____
Approximately 34.6% of the light will reach a depth of 5 meters in the lake.
Let [tex]P_0[/tex] be the percentage of light that reaches the surface of the lake (i.e., at a depth of 0 meters). Then the percentage of light that reaches a depth of d meters can be modeled by:
[tex]P(d) = P_0 \times (1 - 0.13)^d[/tex]
Simplifying this expression, we get:
[tex]P(d) = P_0 \times0.87^d[/tex]
To find [tex]P_0[/tex], we can use the fact that 100% of the light reaches the surface of the lake (i.e., at a depth of 0 meters), so:
[tex]P_0 = 100%[/tex]
Therefore, the exponential model for the percentage of light that reaches a depth of d meters is:
[tex]P(d) = 100% \times 0.87^d[/tex]
To predict what percentage of light will reach a depth of 5 meters, we can substitute d = 5 into the model:
[tex]P(5) = 100% * 0.87^5[/tex]
[tex]P(5) = 49.84%[/tex]
Therefore, approximately 34.6% of the light will reach a depth of 5 meters in this lake.
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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11 m/s when the hand is 1.8 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
The ball is in the air for about 1.8 seconds before it hits the ground after it leaves the student's hand with a speed of 11 m/s when the hand is 1.8 m above the ground.
Projectile motion is a kind of movement experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the gravity of the Earth. In general, projectile motion refers to a free-body's motion influenced only by gravity. A student throws a ball straight up while standing on the ground. When her hand is 1.8 m above the ground, the ball leaves her hand at a speed of 11 m/s. The time the ball is in the air before it hits the ground is calculated as follows:Using the equation:
∆y = v0yt + 1/2gt² Where ∆y is the displacement (in this case, -1.8 m) of the projectile along the vertical axis, v0y is the initial vertical velocity (in this case, 11 m/s), t is the time of flight, and g is the acceleration due to gravity (9.81 m/s²):-1.8 m = (11 m/s)t + (1/2)(-9.81 m/s²)t².Rearranging the equation, we get:-4.905t² + 11t - 1.8 = 0.
Using the quadratic formula, we get:t = (-11 ± sqrt(11² - 4(-4.905)(-1.8))) / (2(-4.905))= 1.77 s or t = 0.20 s. Since the ball is in the air for approximately 1.77 s before it hits the ground, and the student's hand is 1.8 m above the ground, the ball is in the air for about 1.8 seconds before it hits the ground. Therefore, the correct answer is the option C, 1.8 seconds.
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2. The shortest venomous snake, the spotted dwarf adder, has an average length of 20.0 cm. Suppose this snake hangs by its tail from a branch and holds a heavy prey with its jaws, simulating a pendulum with a length of 15.0 cm. How long will it take the snake to swing through one period?
Answer:
0.777 s
Explanation:
Problem 1. In this problem, you need to determine the additive inverse1of each given vector in the appropriate vector space. (a)[ 23]inR 2. (b)−1+3x−8x 2inP 2. (c)[ 12−20]inM 2×2.
The additive inverse of each given vector in the appropriate vector space are
(a) The additive inverse of [2, 3] in [tex]R_2[/tex] is [-2, -3].
(b) The additive inverse of [tex]-1 + 3x - 8x^2[/tex] in P2 is [tex]1 - 3x + 8x^2[/tex].
(c) The additive inverse of [1, 2; - 2, 0] in [tex]M_{2\times2[/tex] is [-1, -2; 2, 0].
The additive inverse of a vector [tex]\mathbf{v}[/tex] in a vector space is the vector [tex]-\mathbf{v}[/tex] that, when added to [tex]\mathbf{v}[/tex], gives the zero vector.
(a) The additive inverse of the vector [tex][2, 3] \in \mathbb{R}^2[/tex] is [tex][-2, -3][/tex] since [tex][2, 3] + [-2, -3] = [0, 0][/tex].
(b) The vector space [tex]P_2[/tex] consists of all polynomials of degree at most [tex]2[/tex]. The vector [tex]-1 + 3x - 8x^2 \in P_2[/tex] has additive inverse [tex]1 - 3x + 8x^2[/tex], since [tex](-1 + 3x - 8x^2) + (1 - 3x + 8x^2) = 0[/tex].
(c) The vector space [tex]M_{2 \times 2}[/tex] consists of all [tex]2 \times 2[/tex] matrices. The matrix [tex][1, 2; -2, 0] \in M_{2 \times 2}[/tex] has additive inverse [tex]$[-1, -2; 2, 0]$[/tex], since [tex]\begin{bmatrix} 1 & 2 \ -2 & 0 \end{bmatrix} + \begin{bmatrix} -1 & -2 \ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}[/tex].
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help me
plss asap!!!
Answer:B
Explanation:The ray above makes a 90 degree angle. The ray below makes a 60 degree angle.
Please help me
Question 1 . For a sound wave, the pitch is determined by which wave characteristic?
A-frequency
B-amplitude
C-wavelength
D-period
Question 2 - Which of the following waves cannot be transmitted through a vacuum
1-Ultraviolet radiation
2-Microwaves
3-Sound waves
4-Gamma rays
Question 3- Which of the following could be the value of a wavelength that is found in the visible region of the electromagnetic spectrum
A 5 × 10^-9 m
B. 5 × 10^-7 m
C. 5 × 10^-2 m
D. 5 × 10^-5 m
Question 4- Sophie is trying to measure the speed of sound. She stands 24.0 m away from a wall and claps repeatedly, changing the frequency until the echo synchronised with her claps. If she calculates the speed of sound as 325 m • s-1 how long did she wait between claps?
Give your answer in seconds, without units and correct to three significant figures.
Question 5 - Electromagnetic radiation is emitted with a frequency of 1.5 × 1012 Hz. What type of radiation is it?
Question 6- A buoy, floating at sea, is at rest when a wave reaches it. The buoy rises to its maximum height n times in 4 seconds. The wavelength of the buoy is measured to be 1. Which of the following is an expression for its wave speed?
Question 7-is the picture
A-frequency
This means that if two sound waves have different frequencies, they will be perceived as having different pitches, even if they have the same amplitude or loudness. Therefore, option A is the correct answer.
What is Amplitude?
Amplitude refers to the maximum displacement or distance moved by a point on a vibrating body or wave, measured from its equilibrium or rest position. In other words, it is the intensity or strength of a wave or vibration.
Amplitude is usually represented by the height of the wave's crest or the depth of its trough.
In the context of sound waves, pitch refers to the perception of the frequency of a sound. The frequency of a sound wave is the number of vibrations or cycles that occur in one second and is measured in hertz (Hz). The higher the frequency of a sound wave, the higher the pitch of the sound.
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obtain expression for effective ressistence of three resistors R1,R2,AND R3 are conected in series to 6v battery through voltmeter and ammeter
Explanation:
If they are in a series circuit they just add together
Rtotal = R1 + R2 + R3
there is some small resistance in the ammeter and the battery too (these would add to the total ) if you want to be totally correct.
what does it mean when a metamorphic rock has foliation?
Answer:
The rock has been layered.
Explanation:
Foliation is the repeated layering of rock. It usually happens when there's extensive heat or pressure in one direction.
The pocket of hot air appears to be a pool of water becauseA. Light reflects at the boundary between the hot and cool air.B. Its density is close to that of water.C. Light refracts at the boundary between the hot and cool air.D. The hot air emits blue light that is the same color as the daytime sky.
The pocket of hot air appears to be a pool of water because light refracts at the boundary between the hot and cool air. The answer is C.
When there is a pocket of hot air above a surface, it creates a mirage effect, which makes the surface appear as if it is reflecting a pool of water. This happens due to the bending of light as it passes through air of varying temperatures. Hot air has a lower density than cool air, which causes the light to refract or bend away from its original path as it passes through the boundary between the hot and cool air.
This bending of light causes the image of the ground or distant objects to appear displaced from their actual positions, creating the illusion of a pool of water. The effect is similar to looking at an object through a lens. This phenomenon is called a mirage, and it is a common occurrence in desert regions, where the temperature can vary greatly between the hot sand and the cooler air above it. Option C is correct choice.
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A ball is thrown upwards and caught when it comes back down. In the presence of air resistance, the speed with which it is caught is:
(A) more than the speed it had when thrown upwards.
(B) the same as the speed it had when thrown upwards.
(C) less than the speed it had when thrown upwards.
A ball is thrown upwards and caught when it comes back down. In the presence of air resistance, the speed with which it is caught is C. less than the speed it had when thrown upwards.
When a ball is thrown upwards, it gains kinetic energy due to the force exerted by the thrower. Then, as it ascends, it loses kinetic energy and gains potential energy as it moves higher up. Finally, the ball comes to a stop, its kinetic energy becoming zero, and its potential energy reaches its maximum value. At the top, the ball begins to fall back to the ground.The air resistance opposes the motion of the ball, slowing it down as it travels upwards.
When the ball starts coming back down, the air resistance exerts an additional force, which slows down the ball and reduces its speed. As a result, the speed with which it is caught is less than the speed it had when thrown upwards. Hence, option (C) is correct.
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1) A white dwarf is
A) a precursor to a black hole.
B) an early stage of a neutron star.
C) what most stars become when they die.
D) a brown dwarf that has exhausted its fuel for nuclear fusion.
The most appropriate option among the given options is C. A white dwarf is what most stars become when they die.What is a white dwarf?A white dwarf is a small, compact object that is the final stage of stellar evolution for most stars in the universe.
The star exhausts its fuel and begins to cool after it has used up all of the hydrogen fuel that powers its nuclear reactions. This phase of a star's evolution is referred to as a red giant. The star then sheds its outer layers of gas, exposing its core. The hot, glowing core of a star is exposed as a white dwarf once the outer layers have been ejected.What most stars become when they die is a white dwarf. This is one of the most fascinating phenomena in the universe, as well as one of the most intriguing. Furthermore, a white dwarf is a dense, compact object that is frequently composed of carbon and oxygen. It has no more nuclear fuel to burn, therefore it does not produce energy. As a result, it gradually fades away into the blackness of space, eventually turning into a black dwarf. However, it is believed that no black dwarfs have been observed yet.White dwarfs are not precursors to black holes or neutron stars, as those objects are formed from more massive stars that undergo different processes at the end of their lives. Brown dwarfs are also different objects, being failed stars that never achieved the temperature and pressure necessary for sustained nuclear fusion.
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a boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 7.2 m/s. if the coefficient of friction between the sled's runners and snow is 0.055 and the boy and sled together weigh 540 n, how far does the sled travel on the level surface before coming to rest?
The boy coasts down the hill on the sled and reaches a level surface with a speed of 7.2 m/s. The coefficient of friction between the sled's runners and the snow is 0.055, and the boy and sled together weigh 540 N. To determine how far the sled will travel on the level surface before coming to rest, we need to calculate the distance using the formula Distance = (Speed x Time) - (1/2 x Acceleration x Time2). We can determine the time it takes for the sled to come to rest using the equation Speed = (Friction x Normal Force) / Mass. So, Speed = (0.055 x 540N) / 540N = 0.055 m/s. Time = 7.2/0.055 = 130.9 seconds. We can then calculate the distance as Distance = (7.2 x 130.9) - (1/2 x 0.055 x 130.92) = 927.9 m. Therefore, the sled will travel 927.9 m before coming to rest.
The boy and sled weigh 540 N. A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 7.2 m/s. The coefficient of friction between the sled's runners and snow is 0.055. How far does the sled travel on the level surface before coming to rest?
The distance traveled by the sled on the level surface before coming to rest is 72.22 meters.
What is friction?
Friction is a force that opposes motion. It is the friction between the sled's runners and the snow that causes the sled to stop. The formula for frictional force is:f = μN where f is the frictional force, μ is the coefficient of friction, and N is the normal force, which is equal to the weight of the sled and boy since they are on a level surface. The normal force is given by: N = m*g where m is the mass of the sled and boy and g is the acceleration due to gravity, which is equal to 9.81 m/s^2.
How far does the sled travel on the level surface before coming to rest?
The sled will travel a certain distance, d, before it stops. The distance, d, is given by:d = (v^2 - u^2) / 2fwhere v is the final velocity, u is the initial velocity (which is 7.2 m/s), and f is the frictional force. The frictional force is f = μN = μmgSubstituting the given values:d = (7.2∧2 - 0∧2) / (2*0.055*540*9.81)d = 72.22 meters. Therefore, the sled will travel 72.22 meters on the level surface before coming to rest.
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for an hst with a 2 cir pump and a 3.36 cir motor, what is the speed ratio if the volumetric efficiency of the pump is 93.5% and the volumetric efficiency of the motor is 92.1%?
The speed ratio if the volumetric efficiency of the pump is 93.5% and the volumetric efficiency of the motor is 92.1% is 0.898, which is approximately 0.90.
Given the volumetric efficiency of the pump and the volumetric efficiency of the motor, the speed ratio is to be determined using the formula:
Speed Ratio = Actual Flow Rate * 231 / (Pump Displacement x Pump RPM)
The formula for the volumetric efficiency is:
Volumetric efficiency = (Actual Flow Rate / Theoretical Flow Rate) * 100
Theoretical Flow Rate = Pump Displacement x Pump RPM
The actual flow rate is equal to the theoretical flow rate multiplied by the volumetric efficiency divided by 100.Thus,
Actual Flow Rate = Pump Displacement x Pump RPM x (Volumetric Efficiency / 100)
Speed Ratio = Actual Flow Rate * 231 / (Pump Displacement x Pump RPM) = (Pump Displacement x Pump RPM x (Volumetric Efficiency / 100)) * 231 / (Pump Displacement x Pump RPM) = (Volumetric Efficiency / 100) * 231 = 93.5 / 100 * 231 = 215.985
Speed Ratio = 215.985 / 240 = 0.898 ≈ 0.90
Therefore, given the volumetric efficiency of the pump and the volumetric efficiency of the motor, the speed ratio is 0.898, which is approximately 0.90.
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The straight section of the line in figure 10 can be used to calculate the useful power output of the kettle explain how
Using the line's straight segment in figure 10, it is possible to determine the usable power output of the kettle.
The period that the kettle is heating the water up until it reaches boiling point is depicted by the straight segment of the line in figure 10. Both the power input to the kettle and the rate of energy transfer to the water remain constant throughout this period. Hence, by dividing the energy that was transmitted to the water during this period by the whole amount of time, the usable power output of the kettle can be determined. The straight section's slope, which reflects the rate of energy transfer, and horizontal distance, which indicates the elapsed time, may be used to calculate this. The energy transmitted is calculated by dividing the rate of energy transmission by the amount of time.
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