All categories

3 years ago

4. Both trains speed up all the time.

Physics

1 answer:

Anit [1.1K]3 years ago

4 0

Answer:

The velocity with which the ball strikes the ground = -5.7 m/s

The velocity with which the ball leaves the ground = 4.3 m/s

Explanation:

Seems like there's parts of a couple questions. I'll focus on the tennis ball part which seems to be a full question.

To find the velocity with which the tennis ball hits the ground, we only need to worry about what happens up to that point. We can ignore the rebound for this part. Given:

d = -1.85

a = -9.8

vi = 0

vf = ?

$v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(-9.8)(-1.65)\\v_f^2=32.34\\v_f=-5.7$

*Keep in mind that the square root gives us two answers, a positve and a negative one. We use the negative one here because the final speed is downwards and the question says down is negative.

To find the velocity with which the ball leaves the ground, we only need to worry about what happens after that point. We can ignore the drop for this part. Given:

d = 0.947

a = -9.8

vf = 0

vi = ?

$v_f^2 = v_i^2 + 2ad\\0=v_i^2+2(-9.8)(0.947)\\v_i^2=18.56\\v_i=4.3$

*Here we use the positive speed after the square root because we know the ball is going up, which the question designates as the positive direction.

You might be interested in

(i) A stone is thrown vertically upwards with an initial velocity of 60ms. Find

elixir [45]

Answer:

plzzzzzzz follow me back

5 0

3 years ago

Read 2 more answers

A star is rotating at 5.55x10-5 π rad/s, its velocity decelerates at a constant rate of 0.5x10-9 π rad/s2.?

Basile [38]

<h2>The current rotational period of that star is 10.01 hours.</h2>

Explanation:

Given that,

Initial angular velocity of the star, $\omega=5.55\times 10^{-5}\pi \ rad/s$

It decelerates, final angular speed, $\omega_f=0$

Deceleration, $\alpha =-0.5\times 10^{-9}\pi \ rad/s^2$

It is not required to use the rotational kinematics formula. The angular velocity in terms of time period is given by :

$\omega=\dfrac{2\pi}{T}$

T is current rotational period of that star

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{5.55\times 10^{-5}\pi \ rad/s}$

T = 36036.03 second

1 hour = 3600 seconds

So, T = 10.01 hours

So, the current rotational period of that star is 10.01 hours. Hence, this is the required solution. Hence, this is the required solution.

4 0

3 years ago

an object is moving in a straight line with a constant acceleration initially it is traveling at 16 meters per second. 3 seconds

VLD [36.1K]

A)It moved 6 meters.

B)It would 3 meters.

I think this is the answer.

Have a good day.

4 0

3 years ago

What is another word that can be used to describe the position of the<br> object?

steposvetlana [31]

where the object is located

6 0

3 years ago

If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move

alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: $m=22kg$

coefficient of friction: $\mu =0.27$

and we also know the acceleration of gravity is $g=9.81m/s^2$

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

weight: $w=mg$ (acting down from)

so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

$ma_{y}=N-mg$

and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

$0=N-mg\\N=mg$

-----------

now let's analyze the horizontal forces

frictional force: $f= \mu N$ and since $N=mg$ --> $f=\mu mg$

force to move the object: $F$

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

$F=ma_{x}=F-f\\ma_{x}=F-\mu mg$

and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

and we get:

$0=F-\mu mg\\F=\mu mg$

substituting known values:

$F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N$

3 0

3 years ago