Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2

6 0

3 years ago

An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to

olga2289 [7]

Answer:

The speed decreases.

Explanation:

This can be explained using the conservation of linear momentum.

Since there is no friction, the initial moment of the train must be equal to its linear moment after it is filled with water.

the initial linear momentum is

$m_{1}v_{1}$

where $m_{1}$ is the initial mass of the train, and $v_{1}$ the initial speed of the train.

And linear momentum after the water filled the train car is

$m_{2}v_{2}$

where $m_{2}$ is mass of the train after the rain, and $v_{2}$ the speed of the train after the rain

<u>the equality must be fulfilled:</u>

$m_{1}v_{1}=m_{2}v_{2}$

We know that if water is added to the train, $m_{2}$ that is the mass after the water is added, is greater than $m_{1}$ which is the mass of the train without the water.

Therefore, in order for the conservation of the linear momentum to be fulfilled: $m_{1}v_{1}=m_{2}v_{2}$

the speed after the water is added ( $v_{2}$ ) must be smaller than the initial train speed ( $v_{1}$ ) . So the speed of the car decreases.

3 0

3 years ago

You drop a stone down off a bridge. You are able to count to 4.0 seconds when it finally hits the water. How high is the bridge?

mart [117]

Answer:

The height of the bridge is 78.4 m.

Explanation:

Given;

time of the stone motion off the bridge, t = 4.0 s

acceleration due to gravity, g = 9.8 m/s²

The height of the bridge is given by;

h = ut + ¹/₂gt²

where;

u is the initial velocity of the stone, u = 0

h = ¹/₂gt²

h = ¹/₂(9.8)(4)²

h = 78.4 m

Therefore, the height of the bridge is 78.4 m.

7 0

3 years ago