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charle [14.2K]
3 years ago
5

Long ding dong what it do?

Physics
2 answers:
Nezavi [6.7K]3 years ago
7 0

Answer:

long very very

Andre45 [30]3 years ago
3 0

Answer:

nun wbu

Explanation:

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If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
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The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

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3 years ago
Which is normally greater, the energy in ordinary sound or the energy in ordinary light? how does the speed of sound compare to
Thepotemich [5.8K]

The energy in ordinary light is greater than the energy in ordinary sound

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Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.

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Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

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