Solar energy can be changed into thermal energy without any work being done. true or false

Anyone who guesses what this means gets brainliest answer!!

hram777 [196]

Answer:

I believe this is German for something day how are you????? I was talking about the holocaust again in school and decided to start learning German so if im wrong then.. wowww!

Explanation:

3 0

3 years ago

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When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

as we know that

$m_1 = 5 kg$

$m_2 = 2 kg$

$v_{1i} = 2.8 m/s$

$v_{2i} = 0 m/s$

now from above equation we have

$5(2.8) + 2(0) = m_1v+ m_2v$

$14 = (5+ 2) v$

$v = 2 m/s$

so the speed of combined system is 2 m/s

8 0

3 years ago

Which acceleration-time graph corresponds to the motion of the car if it moves toward the right, while slowing down at a steady

Bumek [7]

We define acceleration as the rate of change of the velocity

Thus, if you have positive velocity and positive acceleration, your <u>speed increases.</u>

If you have positive velocity and negative acceleration, your speed decreases.

Now you get the idea, we will see that the correct option is graph 1.

We know that the car moves towards the right (let's define this as "the car has positive velocity") and we also know that te car is slowing down constantly (thus the acceleration needs to be negative and constant).

By looking at the graphs, the only one with these properties is graph 1.

If you want to learn more, you can read:

brainly.com/question/12550364

5 0

2 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

4 years ago

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Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?

MrMuchimi

T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,

Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.

</span>

8 0

3 years ago