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Vedmedyk [2.9K]
3 years ago
9

A batter hits a baseball so that it leaves the bat with an initial speed of 60m/s at an initial angle of 42 with the horizontal

Find: The position of the ball and the magnitude and direction of its velocity at t = 2s Find the time when the ball reaches the highest point of its flight Find the height H and the acceleration at this point Find the horizontal range H.
Physics
1 answer:
Elza [17]3 years ago
6 0

Answer: 24.06°

Explanation:

So,we can say after t if it reaches height h then,

h = (37 sin 53)t - 1/2 * 9.8t^2 (as,vertical component of velocity is 37 sin 53)

Given t = 2s

So, h = 39.5m

And horizontal displacement will be

r = 37 cos 53 *2 = 44.52m

So,after 2s the baseball will be lying 39.57m

above its point of projection and 44.52m ahead of its point of projection.

Now.let the vertical component of velocity will become Vy after time 2s

So, Uy = 37 sin 53- 9.8* 2

or, U = 9.95m/s

And.horizontal component of velocity remains

constant i.e Vx = 37 cos 53 = 22.27m/s

So.magnitude of velocity after 2s is

Square root of (Vx^2 + Vy^2)= 24.4m/s

Making an angle of tan 22.27/9.95 = 24.06°

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The ground exerts an equal force on the golf ball
8 0
2 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
3 years ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
1. Describe what is happening at point A when these 3 vectors act on point A. 2. Describe what happen to the resultant vector at
zavuch27 [327]

Answer: Don't know sorry

Explanation:

4 0
2 years ago
Which of the following helps to explain how hydraulic lifts work?
Gnesinka [82]

Answer : C. Pascal's principle.

Explaination : Pascal's principle (well-known as Pascal's law) states that if a closed container contains a fluid at rest, then a small change in pressure at one side of the fluid is transmitted to each and every part of the fluid and also to the walls of the container without any loss. In a hydraulic lift, we need the same mechanism to work and so we take the help of Pascal's principle.


Hence, the correct option is C. Pascal's principle.

8 0
3 years ago
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