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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:

Magnetic field, B=2.39\times 10^{-3}\ T

Explanation:

It is given that,

Number of turns, N = 320

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We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}

B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}

B = 0.00239 T

or

B=2.39\times 10^{-3}\ T

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is 2.39\times 10^{-3}\ T. Hence, this is the required solution.

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A cube of an unknown element has a shiny, silvery color. The side of the cube measures 2.0 cm and the cube has a mass of 14.56 g
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3 years ago
Distances (d)=144m<br> Time (t)=12s<br><br><br> What is the speed?
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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