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3 years ago

What does the SLOPE of a Velocity-Time graph represent? * 1 point

Physics

1 answer:

lyudmila [28]3 years ago

6 0

Acceleration

Explanation:

In the Velocity-Time graph Velocity is taken in Y-Axis and plotted against Time in X-Axis.
The unit of velocity is mentioned in m/s and unit of time is mentioned in seconds.
Here we can analyze the instantaneous acceleration and displacement covered.
From the slope we can get the body accelerated.
From the area under the graph we can get the displacement traveled.
We can calculate both average and Instantaneous acceleration from the slope of this graph.

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Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha weighs 637 N, and each boot has a mass of 4.50 kg. A

Studentka2010 [4]

Answer:

-0.73 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, in absence of external forces (the ice is frictionless, so no friction), the total momentum of Aisha + two boots is conserved.

At the beginning, their total momentum is zero, since they are at rest:

$p_i = 0$ (1)

After, their total momentum is:

$p_f = Mv + 2mv'$ (2)

where:

M is Aisha's mass

v is Aisha's velocity relative to the ground

m = 4.50 kg is the mass of each boot

v' is the boot's velocity relative to the ground

We can find:

$M=\frac{W}{g}=\frac{637 N}{9.8 N/kg}=65 kg$ is Aisha's mass (where W = 637 N was her weight)

v' can be rewritten as:

$v'=v+6$

because 6 m/s is the velocity of the boots relative to her, while v' is their velocity relative to the ground.

Substituting and combining (1) and (2) we find:

$0=Mv+2m(v+6)\\0=Mv+2mv+12m\\v=\frac{-12m}{M+2m}=\frac{-12(4.50)}{65+2(4.50)}=-0.73 m/s$

and the negative sign indicates that the direction is opposite to that of the boots.

8 0

3 years ago

A child threw a stone straight down off a high bridge. The initial velocity of the stone was 15.0 m/s (include units and proper

Andre45 [30]

Answer:

The speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.

Explanation:

Given that, a child threw a stone straight down off a high bridge.

Initial velocity of the stone, u = 15 m/s

We need to find the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant. When it come down, it is moving under the action of gravity. Using equation of motion as :

$v=u+gt\\\\v=15+10\times 3\\\\v=45\ m/s$

So, the speed of the stone before it hit the river 3.00 sec later. Let v is the velocity at that instant is 45 m/s.

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OverLord2011 [107]

The energy stays the same

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katovenus [111]

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