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Alexeev081 [22]
3 years ago
13

A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?

Physics
2 answers:
alexandr1967 [171]3 years ago
7 0

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

v=u+at

Here, train starts from rest so, u = 0

v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

Leni [432]3 years ago
7 0

Answer:

7.5m/s

Explanation:

We use the formula to find the final velocity for the <u>movement with constant acceleration:</u>

V_{f}=V_{i}+at

where

V_{f} is the final velocity of the train

V_{i} is the initial velocity of the train, since it starts from rest V_{i}=0

a is the acceleration, a=1.5m/s^2

and t is the interval of time,  t=5s

So, replacing the values in the formula we get:

V_{f}=0+(1.5m/s^2)(5s)

V_{f}=7.5m/s

The final velocity of the train is   7.5m/s

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A balloon filled with helium gas at 1.00 atm occupies 15.6 L. Will the volume of the balloon increase or decrease in the upper a
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Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

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\mathrm{F}=\mathrm{m} \times \mathrm{a}

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Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

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F = 4.5 N

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