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kaheart [24]
3 years ago
14

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters: what is the correspon

ding time interval?
A. 2 seconds
B. 4 seconds
C. 5 seconds
D. 10 seconds
E. 15 seconds

Physics
1 answer:
shtirl [24]3 years ago
6 0

Answer: The correct answer is option (D).

Explanation:

The area under the line is 10 meters that is displacement which means that area enclosed by the right angled triangle is 10 unit squares.

So, the it can be written as:

\text{area of triangle}=\frac{1}{2}\times velocity\times time

10=\frac{1}{2}\times 2.0 m/s\times time

time =10 seconds

Hence the, the correct answer is option (D).

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now we will have

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d = 2.4 m

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now here we have

\frac{\lambda}{2} = \frac{0.20\times 2.4}{10}

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In which scenario will the two objects have the least gravitational force between them?​
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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
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Question: How fast was the arrow moving before it joined the block?

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The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

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m_b = 1.40kg,

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and simplify to get:

\boxed{ v= 15.9m/s}

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3 years ago
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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
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T^2=\dfrac{4\pi^2}{GM}r^3

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