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3 years ago

A train moves with a constant velocity for 15s for 155m. How fast is the train moving?

Physics

1 answer:

bonufazy [111]3 years ago

5 0

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1): distance = velocity times time <=> d = v*t for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2): velocity = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

v = 155 / 15 <=> v = 10.3

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Which statement best describes the relationship between plants and oxygen?

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A statement would be to the effect that plants use oxygen as part of photosynthesis

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4 years ago

Read 2 more answers

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\\h=19.62 meters$

Part b

$v=u+at\\v=0+9.81*2\\v=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs$

4 0

3 years ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

ankoles [38]

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
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so W=12*100
W=1200J

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3 years ago

Explain how surface and subsurface events are integral parts of the rock cycle.

vagabundo [1.1K]

Example of surface events are erosion and weathering. Erosion is the carrying of a particle from one place to the other and weathering is the breaking down of particles. These processes help in rock formation because this allows physical changes (grouping together or breaking down) on a certain substance. Subsurface events are those which happened underground such as the flow of underground water which subsequently allow the deposition of minerals, etc.

3 0

4 years ago

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0

3 years ago