Which refers to the distance between two crest or to drafts on a transverse wave

A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t

zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0

3 years ago

The velocity of a moving body increases from 10m/s to 15 m/s in 5 sec . calculate its acceleration

Gnoma [55]

Answer:

v=u+at

15=10+a x 5

15-10= 5a

5= 5a

a=1 m/sec^2

4 0

3 years ago

Which of the following is not true about the Law of Conservation of Energy?

SVEN [57.7K]

By definition, the law of conservation of energy states that:

Ei = Ef

Where,

Ei: initial energy

Ef: final energy

Therefore, no matter the type of energy, always the final energy is equal to the final energy.

Energy can be transformed into another type of energy. For example, the potential energy can be transformed into kinetic energy.

Also, energy is not created, nor destroyed.

Answer:

The following is not true about the Law of Conservation of Energy:

A. It states that the total energy in the universe keeps increasing.

8 0

3 years ago

Read 2 more answers

MULTIPLE CHOICE QUESTION

Nookie1986 [14]

Which object? More information is needed to answer this question

4 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by $\dfrac{v_i+v_f}2$ , and we know that $v_f=-v_i$ .

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago