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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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Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
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Answer:

D/H =15

Explanation:

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  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

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        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
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