The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.
Someone please help me !!
Answer:25
Explanation: because higher means less kinetic energ
How much momentum, in the x-direction, was transferred to the more massive cart, in kilogram meters per second
The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.
Momentum transfered to the more massive cartThe momentum transfered to the more massive cart is determined by applying the principle of conservation of linear momentum as shown below;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
m₁ is the mass of the smaller cartu₁ is the initial velocity of the samller cartm₂ is the mass of the bigger cart = 3m₁u₂ is the initial velocity of the bigger cartv₁ is the final velocity of the smaller cartv₂ is the final veocity of the bigger cart⁻ΔP₁ = ΔP₂
ΔP₂ = m₂v₂ - m₂u₂
ΔP₂ = m₂(v₂ - u₂)
ΔP₂ = 3m₁(v₂ - u₂)
ΔP₂ = 3 x 3.8 x (1.7 - 0)
ΔP₂ = 19.38 kgm/s
Thus, the momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.
The complete question is beblow
A cart of mass 3.8 kg is traveling to the right (which we will take to be the positive x-direction for this problem) at a speed of 6.9 m/s. It collides with a stationary cart that is three times as massive. After the collision, the more massive cart is moving at a speed of 1.7 m/s, to the right.
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The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. Which of the following statements correctly describes the relationship between m1 and m2 and provides evidence from the graphs?
Answer:
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1 describes the displacement
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
What is the graph represents?The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Therefore, M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
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A magnet gets demagnetized when it is heated.
Answer:
The delicate balance between temperature and magnetic domains is destabilized when a magnet is subjected to high temperatures. If a magnet is exposed to this temperature for an extended length of time or heated over its Curie temperature, it will lose its magnetism and become irreversibly demagnetized.
Explanation:
how many types air
I need a formula of calculating the s i unti of force
The si unit of force is newton.
so, F is eqal to m*g
Answer:
F=ma
Explanation:
force is actually a derived quantity
newton is a unit derived from kg.m/s^2
This is from Newton's second law of motion
F=ma
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of launch with respect to the horizontal? (Assume a flat and horizontal landscape.)
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
the radius of a ball is increasing at a rate of 2 mm per second. how fast is the volume of the ball increasing when the diameter is 40 mm
Step 1: Define an equation that relates the volume of a sphere to its radius.
V = 4/3*π*r3
Step 2: Take the derivative of each side with respect to time (we will define time as "t").
(d/dt)V = (d/dt)(4/3*π*r3)
dV/dt = 4πr2*dr/dt
Step 3: We are told in the problem statement that diameter is 100m, so therefore r = 50mm. We are also told the radius of the sphere is increasing at a rate of 2mm/s, so therefore dr/dt = 2mm/s. We are looking for how fast the volume of the sphere is increasing, or dV/dt.
dV/dt = 4π(50mm)2*(2mm/s)
dV/dt = 62,832 mm3/s
A 7 kg ball of clay traveling at 12 m/s collides with a 25 kg ball of clay traveling in the
same direction at 6 m/s. What is their combined speed if the two balls stick together
when they touch?
Answer:
Given:
m1 = 7 kg
V1 = 12 m/s
m2 = 25 kg
V2 = 6 m/s
To find:
Combined speed of two balls stick together after collision V = ?
Solution:
According to law of conservation of momentum,
m1V1 + m2V2 = (m1+m2)V
7×12 + 25×6 = (7+25)V
84 + 150 = 32V
V = 234/32
V = 7.31 m/s
Combined speed of two ball is 7.31 m/s
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478 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to compress the gas by a factor of 12.0, starting from its initial volume?
I was thinking of using PV = W formula. Like
478 = P(V/2)
956 = PV
W = P(V/12)
12W/V = P
956 = (12W/V)(V)
956 = 12W
W = 79.66667 J
is this correct? could someone please help?
Answer:
Explanation:
I don't think so. Think about it. To compress the volume by a factor of 2 it takes 956 Joules.
Now you come along and you want to get the pressure for 1/12 of the volume. It's going to take a huge pressure to do that.
I would suggest that you have to use a modified form of the formula.
PV = 956
You need to compress the volume by 1/6
P(V/6) = 956
6 * PV/6 = 6 * 956
PV = 5736 J
Why did I only take 1/6? Because. 956 represents the pressure needed for 1/2 the volume. You need to multiply 1/2 * 1/6 to get 1/12
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
When you apply increasing thermal energy to a certain material, it reaches a temperature of 50 degrees C. However, when it reaches this temperature, applying more increasing thermal energy does not cause the temperature to rise. Which of the following best explains what is happening?
1. The system is having its heat leaking out that is not going into the material
2. The material is at its phase change temperature and the thermal energy is going to change the phase instead of increase the temperature
3. The material is made of a heat resistant alloy that is preventing the thermal energy from being fully absorbed
4. The material has already absorbed enough thermal energy and cannot absorb any more
The material is at its phase change temperature and the thermal energy is going to change the phase instead of increase the temperature.
What is latent heat?Latent heat is defined as the thermal energy absorbed or released during a phase change of a substance.
Latent heat can also be explained as the energy in hidden form which is supplied or extracted to change the state of a substance without changing its temperature.
From the above explanation, we can conclude that, the material is at its phase change temperature and the thermal energy is going to change the phase instead of increase the temperature.
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What are the three symbols used in Ohm's law. Explain what each symbol represents and give the units for each of the variables.
Answer:
Step by step explanation:
What is the angular momentum at a radius of 2 m with an object of 5 kg at a
velocity of 20 m/s?
The angular momentum is 200 kg m^2 s^-1
what is angular momentum?Angular momentum is the product of linear momentum and the perpendicular distance. Linear momentum is the product of mass and velocity, where radius is the perpendicular distance
Angular momentum = mass * velocity * radius
Angular momentum = 5 * 2 * 20
Angular momentum = 200 kg m^2 s^-1
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When a penny is dropped, it takes 16 seconds. What is its height
A small block, with a mass of 250 g, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, around the loop, and then onto the final level section on the right. The maximum height of the incline is 80 cm, and the radius of the loop is 15 cm.
a.) Find the initial energy of the block.
b.) Find the velocity of the block at the bottom of the loop.
c.) Find the velocity of the block at the top of the loop.
(a) The initial energy of the block due to its position is 1.96 J.
(b) The velocity of the block at the bottom of the loop is 3.96 m/s.
(c) the velocity of the block at the top of the loop is 3.13 m/s.
Initial energy of the blockThe initial energy of the block due to its position is calculated as follows;
P.E = mgh
P.E = 0.25 X 9.8 X 0.8
P.E = 1.96 J
Conversation of the energyThe velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;
P.Ei + P.Ef = K.Ei + K.Ef
1.96 + 0 = 0 + ¹/₂mvf²
vf² = 2(1.96)/m
vf² = (2 x 1.96) / (0.25)
vf² = 15.68
vf = √15.68
vf = 3.96 m/s
Velocity of the block at top of the loopThe velocity of the block at the top is calculated by applying principle of conservation of energy,
P.Ei + P.Ef = K.Ei + K.Ef
1.96 = mghf + ¹/₂mvf²
where;
hf is the position of the ball at the top of the loop = 2r = 2 x 15 cm = 30 cm = 0.31.96 = 0.25 x 9.8 x 0.3 + 0.5 x 0.25vf²
1.225 = 0.125vf²
vf² = 1.225/0.125
vf² = 9.8
vf = 3.13 m/s
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explain how a deflection magnetometer can be used to find the horizontal component of the Earth's magnetic field
Alex (31kg) and Cassie (19Kg) sit on a 10kg metre-long see-saw at the local park. The pivot of the see-saw is in the middle of its length. If Cassie sits at one end of the see-saw, where relative to the other end must Alex sit so the net torque is balanced? (unit:metres)
Answer:
M1 g L1 = 19 kg * 9.8 m/s^2 * 5 m = counter clockwise torque - Cassie at left end
M1 g L1 = M2 g L2 for torques to balance
L2 = M1 L1 / M2 = 19 * 5 / 31 = 3.06 M
Alex should sit at 3.1 m from the fulcrum (at 5 m from each end)
Which planet is the farthest?
Which statement best describes the circular flow model?
2. All of the following are examples of physical properties except:
A. tearing B. density C. melting point D. boiling point
All of the following are examples of physical properties except tearing.
What is Physical property?
This is used to describe the state of a physical system and is usually measurable.
Examples include:
DensityMelting point Boiling pointTearing isn't an example of a physical property which was why option A was chosen as the most appropriate choice.
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What type of heat transfer occurs in your stomach when you eat hot soup and an ice cold beverage
The type of heat transfer occurs in your stomach when you eat hot soup and an ice cold beverage is Conduction.
What is Conduction?This is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules.
This happens when they are in close contact with each other which was why Conduction was chosen as the most appropriate choice.
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Which of the following is an example of the characteristic of excretion?
A) We shiver when we get cold.
B) Moss on the side of the tree is active even though it looks still.
C) Human kidneys produce urine.
D) A rabbit gets nutrients from a carrot.
Answer:
C
because urine is waste product
A car of mass 1000 kg moves 3 km east in a straight line and then 4 km north. What is the total distance and displacement of the car from the initial position?
The net (resultant) force on the car is
Select one:
a) distance = 7 km and displacement = 5 km
b) distance = 5 km and displacement =7 km.
c) distance = 25 km and displacement =7 km.
d) distance = 7 km and displacement = 25 km
Answer:
a
Explanation:
Distance is simply the distance travelled which in this case would be 4km + 3km = 7km
To work out displacement, try to imagine the situation.
Draw a straight line to the east (label it 3) and then draw another line from the end of the first line upwards (label this one 4). Thus, you've created a right angles triangle. Now use pythagorean theorem to work out the displacement
4^2 + 3^2 = 25
sqrt 25 = 5 = displacement
What is the force of the drag for a 65 kg bicyclist, initially at rest at the top of a hill coasts down the hill, reaching a speed of 15.5 m/s at the bottom of the hill. The distance is 60M. neglect any friction impeding the motion and the rotational energy of the wheels.
Height is 19M
Intial GPE is 12350J
KE is 7808J and loss is 4542 J
When resting, a person generates about 412005 joules of heat from the body. The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C. If the heat from the person goes only into the water, find the water temperature.
If a person generates about 412005 joules of heat from the body, the water temperature is mathematically given as
t=21.6296C
What is the water temperature.?Question Parameter(s):
The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C
Generally, the equation for the Heat is mathematically given as
Heat gained =Heat loess
Thereofore
mw*cw*(t-2160)=1.5*10^5
[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]
t=21.6296C
In conclusion, the tempreature
t=21.6296C
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How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 20 N/m?
[tex]\\ \rm\rightarrowtail F=-kx[/tex]
[tex]\\ \rm\rightarrowtail F=-20(0.03)[/tex]
[tex]\\ \rm\rightarrowtail F=-0.6N[/tex]
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 39.2 m
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.
Which of the following particles is similar to a He nucleus?
alpha
beta
gamma
neutrino
Alpha
I hope this helps you
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A bucket of mass m is attached to a rope that is wound around the outside of a solid sphere (I = 2/5 M^2) of radius R. When the bucket is allowed to fall from rest, it falls with an acceleration of a down. What is the mass of the sphere in terms of m, R, a, and g?
Answer:
[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)
Explanation:
The forces on the bucket are:
Weight of the bucket: [tex]m\, g[/tex] (downward.)Tension in the rope (upward.)Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:
[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].
The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].
Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:
[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].
At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].
Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].
Therefore, the magnitude of the angular acceleration of this sphere would be:
[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].
The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].
Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].
Hence the equation:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].
Solve this equation for [tex]M[/tex], the mass of this sphere:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].
[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].
[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].
You see a car that appears very small, so you assume that it must be far from you. You are using the monocular cue of
The monocular cue of relative size