A 15-cm-focal-length converging lens is 19 cm to the right of a 6.0-cm-focal-length converging lens. A 1.0-cm-tall object is distance L to the left of the 6.0-cm-focal-length lens.
Required:
For what value of L is the final image of this two-lens system halfway between the two lenses?
Answer:
L = 11.014 cm
Explanation:
Halfway between the two lenses is 19/2 = 9.5 cm.
Thus, this means virtually with respect to lens, the final image is at -9.5 cm
Thus, from here, we will work this out backwards.
Let's first solve for the initial position of the object for the second lens;
(1/S2) + (1/s'2) = (1/f2)
Where s'2 is the real image.
F2 is focal length
Thus;
(1/s'2) = (1/f2) - (1/s2)
(1/s'2) = (1/15) - (1/-9.5)
(1/s'2) = 0.1719
s'2 = 5.82 cm
The object for the second lens is located at 5.82 cm in front of the second lens.
Now, The object for the second lens and the image for the first lens will be the same.
This means the distance of the image from the first lens is at; 19 - 5.82 = 13.18 cm.
Now let's solve for the object distance of the first lens which will be denoted by L.
1/L = (1/f1) + (1/s'1)
Where f1 = 6 cm
1/L = (1/6) - (1/13.18)
1/L = 0.090794
L = 1/0.090794
L = 11.014 cm
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
What must a scientist do in order to develop a testable hypothesis?
A. Identify a conclusion that provides the most popular explanation
for the question.
B. Determine whether experimental observations can provide
evidence to support a conclusion.
C. Copy the opinions of other scientists with similar questions.
D. Survey the preferences of other scientists who have done similar
research.
The correct answer is option B. Determine whether experimental observations can provide evidence to support a conclusion.
1. Identify the research question: Begin by clearly defining the question or problem that you want to investigate. This question should be specific and focused, allowing for clear hypotheses to be developed.
2. Conduct background research: Familiarize yourself with existing knowledge and previous research related to your question. This step will help you understand the current state of knowledge in the field and identify any gaps or areas that require further investigation.
3. Formulate a hypothesis: Based on your background research, develop a hypothesis that proposes a possible explanation or answer to your research question. A hypothesis should be a clear statement that can be tested through experimentation or observation. It should be specific, measurable, and falsifiable, meaning that it can be proven wrong if the evidence does not support it.
4. Design experiments or observations: Once you have formulated a hypothesis, consider the experimental or observational methods that can be used to test it. Determine what data you need to collect, what variables you need to manipulate or measure, and what controls should be put in place to ensure the validity of your results.
5. Predict outcomes: Based on your hypothesis, make predictions about the expected outcomes of your experiments or observations. These predictions should be derived from your hypothesis and should be specific enough to be tested.
6. Conduct the experiment or observation: Carry out the planned experiments or observations, ensuring that you collect and record data accurately. Implement proper controls and procedures to minimize any biases or confounding factors that could affect the results.
7. Analyze the data: Once you have collected the data, analyze it using appropriate statistical or analytical methods. Evaluate whether the data supports or contradicts your hypothesis and predictions.
8. Draw conclusions: Based on the analysis of your data, draw conclusions about whether the evidence supports or refutes your hypothesis. Clearly state the findings and discuss their implications in the context of the research question.
9. Communicate the results: Share your findings with the scientific community through scientific papers, presentations, or other appropriate means. Allow other scientists to review and replicate your work, fostering further discussion and advancement in the field.
Remember, the scientific process is iterative, and developing a hypothesis is just the starting point. Scientists continuously refine and revise their hypotheses based on new evidence and insights gained from their experiments and observations.
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A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m.
What is the mass of the cable?
Answer:
m_cable = 2,676 kg
Explanation:
For this exercise we must look for the acceleration with the kinematic ce relations
v² = v₀² + 2 a x
since the block starts from rest, its initial velocity is vo = 0
a = v² / 2x
a = 4.2² /(2 2.0)
a = 4.41 m / s²
now we can use Newton's second law
Note that the mass that the extreme force has to drag is the mass of the block plus the mass of the cable.
F = (m + m_cable) a
m_cable = F / a -m
m_cable = 100 / 4.41 - 20
m_cable = 2,676 kg
4. Brandon throws a tennis ball vertically upward. The ball returns to the point of release after 4.0 s. What is the
speed of the ball as it is released?
Answer:
-39.2m/s
Explanation:
Using the equation of motion;
v = u + at
Since the ball is thrown upward, the acceleration due to gravity acting on it will be negative, hence a = -g
v = u - gt
Since g = 9.8m/s²
t = 4.0s
u = 0m/s
v = 0 + (-9.8)(4)
v = 0 + (-9.8)(4)
v = -39.2m/s
Hence the speed of the ball before release is -39.2m/s
Imagine that you are standing on a spherical asteroid deep in space far from other objects. You pick up a small rock and throw it straight up from the surface of the asteroid. The asteroid has a radius of 9 m and the rock you threw has a mass of 0.113 kg. You notice that if you throw the rock with a velocity less than 45.7 m/s it eventually comes crashing back into the asteroid.
Required:
Calculate the mass of the asteroid.
Answer:
M = 1.409 10¹⁴ kg
Explanation:
In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.
Starting point. When you drop the stone
Em₀ = K + U
Em₀ = ½ m v² - G m M / r
where M and r are the mass and radius of the asteroid
Final point. When the stone is too far from the asteroid
Em_f = U = - G m M / R_f
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v² - G m M / r = - G m M / R_f
½ v² = G M (1 / r - 1 /R_f)
indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞
½ v² = G M (1 /r)
M = [tex]\frac{v^2 r}{2G}[/tex]
let's calculate
M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]
M = 1.409 10¹⁴ kg
Nick and Chloe left their campsite by canoe and paddle downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average speed of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream
The time spent by the campers when they turn around downstream is 15 minutes.
Total distance traveled by Nick and Chloe
The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.
Let time for downstream = t1
Let time for upstream = t2
distance covered in upstream = distance covered in downstream = d
12(t1) = d
4(t2) = d
12t1 = 4t2
t1 + t2 = 1
t2 = 1 - t1
12t1 = 4(1 - t1)
12t1 = 4 - 4t1
16t1 = 4
t1 = 4/16
t1 = 0.25 hours
t1 = 0.25(60 min) = 15 mins
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write two units for electric field intensity and show thier equivalence?
Answer:
Electric field intensity is the force experienced by a test charge q in a electric field E.
Answer:
hope it helps...
Explanation:
The SI unit of the electric field is volts per meter (V/m). This unit is equivalent to Newton's per coulomb. These are derived units where Newton is a unit of force and Coulomb is the unit of charge
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component F _ { x } = - [ 20.0 N + ( 3.0 N / m ) x ]F x =−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
[tex]W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J[/tex]
if a cyclist is travelling a road due east at 12km/h and a wind is blowing from south-west at 5km/h. find the velocity of the wind relative to the cyclist.
Answer:
The velocity of wind with respect to cyclist is [tex]-15.5 \widehat{i} - 3.5 \widehat{j}[/tex].
Explanation:
speed of cyclist = 12 km/h east
speed of wind = 5 km/h south west
Write the speeds in the vector form
[tex]\overrightarrow{vc} = 12 \widehat{i}\\\\overrightarrow{vw} = - 5 (cos 45 \widehat{i} + sin 45 \widehat{j})\\\\\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}[/tex]
The velocity of wind with respect to cyclist is
[tex]\overrightarrow{vw} =-3.5 \widehat{i} - 3.5 \widehat{j}\overrightarrow{v_(w/c)} = \overrightarrow{vw}-\overrightarrow{vc}\\\\\overrightarrow{v_(w/c)} = - 3.5 \widehat{i} - 3.5 \widehat{j} - 12 \widehat{i}\\\\\overrightarrow{v_(w/c)} =-15.5 \widehat{i} - 3.5 \widehat{j}[/tex]
A 2.2 kg, 20-cm-diameter turntable rotates at 80 rpm on frictionless bearings. Two 600 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?
Answer:
[tex]w_2=38.3rpm[/tex]
Explanation:
From the question we are told that:
Mass of turntable [tex]M=2.2kg[/tex]
Diameter of turntable [tex]d=20cm=>0.2m[/tex]
Angular Velocity [tex]\omega =80rpm[/tex]
Mass of Blocks [tex]M_b=600g=>0.6kg[/tex]
Generally the equation for inertia is mathematically given by
Initial scenario at \omega=80rpm
[tex]I_1=\frac{1}{2}mR^2[/tex]
[tex]I_1=\frac{1}{2}*2.2*0.1^2[/tex]
[tex]I_1=0.11kgm^2[/tex]
Final scenario
[tex]I_2=I_1+2mR^2[/tex]
[tex]I_2=0.011+(2*0.6*0.12)[/tex]
[tex]I_2=0.023[/tex]
Generally the equation for The relationship between Angular velocity and inertia is mathematically given by
[tex]I_1w_1=I_2w_2[/tex]
[tex]w_2=\frac{I_1 \omega}{I_2}[/tex]
[tex]w_2=\frac{0.011*80}{0.023}[/tex]
[tex]w_2=38.3rpm[/tex]
chứng minh V=kq/r từ mối liên hệ giữ E và V
Answer:V=Aq=KQr
Explanation:
Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?
Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q
Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.
- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2
Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:
dA=dF.dx=KQ.qx2dx
Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:
A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr
Theo đúng định nghĩa: V=Aq=KQr
A single loop of wire with an area of 0.0900 m^2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.160 T/s.
Reqiured:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) [tex]E=0.0144[/tex]
b) [tex]I=0.024A[/tex]
Explanation:
From the question we are told that:
Area [tex]A=0.09m^2[/tex]
Magnetic Field [tex]B=3.80T[/tex]
Rate [tex]\frac{dB}{dt}=0.160T/s[/tex]
Generally the equation for EmF E is mathematically given by
[tex]E=-A\frac{dB}{dt}[/tex]
[tex]E=-(0.0900*0.160)[/tex]
[tex]E=0.0144[/tex]
b)
at Resistance R=0.60
Generally the equation for Current I is mathematically given by
[tex]E=IR[/tex]
[tex]I=\frac{0.0144}{0.600}[/tex]
[tex]I=0.024A[/tex]
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 7950 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.8 m^3 to 19.4 m^3.
Required:
a. Calculate the work done by the gas. Express your answer with the appropriate units.
b. Calculate the change in internal energy of the gas, Express your answer with the appropriate units.
Answer:
Explanation:
From the question we are told that:
Energy [tex]Q=7950kcal=3.3*10^7[/tex]
Initial Volume [tex]V_1=12.8 m^2[/tex]
Final Volume [tex]V_2=19.4 m^2[/tex]
a)
Generally the equation for Work done is mathematically given by
[tex]W=P \triangle V[/tex]
Where
[tex]P= Pressure at 1atm[/tex]
Therefore
[tex]W=(1.01*10^5)(19.4-12.8)[/tex]
[tex]W=6.67*10^5J[/tex]
a)
Generally the equation for Change in internal energy of the gas is mathematically given by
[tex]\triangle U=Q-W[/tex]
[tex]\triangle U=3.3*10^7-6.67*10^5J[/tex]
[tex]\triangle U=3.2*10^7J[/tex]
A moving object is in equilibrium. What bests describes the motion of the object if mo force changes
•It will change directions
•It will slow down
•It will maintain the state of motion
•It will speed up and ten slow down
Answer:
telekinesis moving things with your mind
a horizontal turntable with radius 0.8 m is rotation about a vertical axis at ts center. a small block sits on the turntable at a distance of 0.4 m from the acis and rotates in a circular path at a speed of 0,800 m/s. what minim coeeficent of static friction between the blok and the turntable is required for the block not to slip
Answer: 0.04
Explanation:
Given
Radius of turntable [tex]r=0.8\ m[/tex]
Block is present at a distance of [tex]r_o=0.4\ m[/tex]
Turntable rotates at a speed of [tex]v=0.8\ m/s[/tex]
angular speed of turntable [tex]\omega =\dfrac{v}{r}[/tex]
[tex]\Rightarrow \omega =\dfrac{0.8}{0.8}\\\Rightarrow \omega =1\ rad/s[/tex]
block will experience a force i.e. centripetal force equal to [tex]m\omega ^2r[/tex]. This must balance the friction force [tex]\mu mg[/tex]
[tex]\Rightarrow m\omega^2 r_o=\mu mg\\\Rightarrow 1^2\times 0.4=\mu \times 9.8\\\Rightarrow \mu =0.04[/tex]
Thus, the coefficient of static friction force is [tex]0.04[/tex]
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of charge from one sphere to the other? Answer in Joules.
Answer:
[tex]W=2.76\times 10^{-6}\ J[/tex]
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
[tex]U=\dfrac{Q^2}{2C}[/tex]
Put all the values,
[tex]U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J[/tex]
So, the work done is [tex]2.76\times 10^{-6}\ J[/tex].
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
Answer:
Option C. 50 N
Explanation:
From the question given above, the following data were obtained:
Force (F) applied = 100 N
Angle (θ) = 60°
Horizontal component (Hₓ) =?
The horizontal component of the force can be obtained as follow:
Hₓ = F × Cos θ
Hₓ = 100 × Cos 60
Hₓ = 100 × 0.5
Hₓ = 50 N
Therefore, the horizontal component of the force acting on the dog is 50 N
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm. a) What is the wave? b) Wrote the wave function, y(x,t) c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
What is the answer bro idont now
calculate the work function that requires a 455 nm photon to eject an electron of a value 0.73 eV
Answer:
W = 2 eV
Explanation:
Given that,
The wavelength of a photon = 455 nm
The kinetic energy of a photon, K = 0.73 eV
We need to find the work function of the electron. It can be solved using Einstein's equation such that,
[tex]W=E-K[/tex]
E is the energy of the photon
So,
[tex]W=\dfrac{hc}{\lambda}-K\\\\W=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{455\times 10^{-9}\times 1.6\times 10^{-19}}-0.73\\\\W= 2.73\ eV-0.73\ eV\\\\W=2\ eV[/tex]
So, the work function of the metal is 2 eV.
Why do you think we see the sun moves across the sky?
answer it for brainlliest
Answer:
From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.
Choose the CORRECT statements. A standing wave is resulted from the superposition of
two waves in such a way both waves:
I. have the same direction.
II. are opposite in direction.
III. have the same frequency.
IV. have different frequency.
V. have the same amplitude.
VI. have different amplitude
A. I, III and V
B. II. IV and VI
C. I, IV and V
D. II, III and V
E I and III
F. Ill and V
Answer:
The answer is D
How is wind generated?
O A. Air molecules move from areas of low pressure to areas of high
pressure.
O B. Air molecules move more slowly where the temperature is higher
and the pressure is lower.
C. Air molecules move more quickly where the temperature is lower
and the pressure is higher.
D. Air molecules move from areas of high pressure to areas of low
pressure.
Answer:
Explanation:
Wind is caused by the uneven heating of the atmosphere by the sun, variations in the earth's surface, and rotation of the earth. ... Wind turbines convert the energy in wind to electricity by rotating propeller-like blades around a rotor. The rotor turns the drive shaft, which turns an electric generator
What must be the same for any two resistors that are connected in a series
If a rock displaces 7 ml of water, what is the volume of the rock?
Answer:
if i am not mistaken the volume is 7, because it only took that much space
3.1Chất điểm chuyển động thẳng với phương trình: x = – 1 + 3t2
– 2t
3
(hệ SI, với t ≥ 0). Chất điểm dừng lại để
đổi chiều chuyển động tại vị trí có tọa độ:
Answer:
eqcubuohwehuuc
Explanation:
the the the the the the the the the the the the the the the the dhueirrjhrhjrirjrheh3jeiiruj fnrhfjjjrjrj fjfiirrjejrjejej jrkrjrjrjrjrjdjdjfjrhruriruru rjridjhwjjsjd
Use a scientific calculator to perform the operation below.
5.92 x 107 + 2.11 x 106
A. 6.13 x 107
B. 2.81 x 101
C. 1.25 x 1014
D. 5.71 x 107
SUBMIT
Answer:
A. 6.13 x 107
Explanation:
Given the expression
5.92 x 10^7 + 2.11 x 10^6
First, we need to convert to the whole number
5.92 x 10^7 = 59200000
2.11 x 10^6 = 2110000
Add both values
59200000 + 2110000
= 61,310,000
Express in standard form
= 6.13 * 10^7
This gives the required result
The more bonds an atom can make, the more likely it is to combine with other aton
Which element is most likely able to make the greatest variety of bonds?
nitrogen
hydrogen
oxygen
carbon
Answer:
carbon has tetra valency while oxygen can make 2 bonds while nitrogen can make 3 bonds and hydrogen can make 1 bond so i think answer is definatly carbon
Explanation:
Aluminium has a specific heat capacity of 900 J/(kg°C).
The internal energy of a 2.0kg block of aluminium increases by 13500 j.
By how much does the temperature of the block increase?
B 0.13°C
C 7.5°C
A
0.067 °C
D 15°C
Answer:
24300000K
Explanation:
Q=mc(change in temperature)
900X2.0=1800.0
1800.0X13500= 24300K
According to specific heat capacity, the temperature of the block increases by 7.5°C.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT
Substitute values in above formula ,Δt=Q/mc=13500/2×900=7.5°C.
Thus, the temperature of the block increases by 7.5°C.
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