- The Grayprational Force<br>with which the earth aracts<br>а,

Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the hori

lys-0071 [83]

Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft

Explanation:

1. use the position (x) equation in parobolic movement to find the time (t)

565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t

t= 3.92 s

2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.

y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - $\frac{32.2 ft/s^{2} *3.92 s^{2} }{2}$

y= 148.32 ft

3. finally add the 3 ft that exist between the home plate and the ball

148.32 ft + 3 ft = 151.32

6 0

3 years ago

A ball of gas becomes a ____ when nuclear fusion begins in its core.

Ira Lisetskai [31]

The answer to this is Protostar.

This is a process where it is gathering mass from its parent molecular cloud. Its a very young star meaning, the star was now born.

Hope this helped :)
Have a great day

8 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$