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uysha [10]
3 years ago
6

The surface of an incline is coated with an experimental substance that is intended to reduce the frictional force between a blo

ck and the
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo

Physics
2 answers:
ankoles [38]3 years ago
6 0

Answer:

b

Explanation:

nikklg [1K]3 years ago
4 0

Answer:

粗糙錯愕額外此時自由KK預約為我

Explanation:

因為嗚嗚嗚粗糙此次四月蘇澳碩果僅存有一位我也在鶯鶯燕燕

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For elastic collision

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s

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8 0
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A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
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So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
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And since 1 horsepower is equal to 746 W, we can rewrite the power as
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