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Triss [41]
2 years ago
7

When the batter hit the foul ball, the baseball moved upward to a delighted fan in the top deck.

Physics
1 answer:
oee [108]2 years ago
6 0

Answer:

B) Kinetic energy increases, potential energy decreases

Explanation:

In a given system, when a body is at rest, v =0m/s, the kinetic energy is at zero while the potential energy is at maximum. However, when a body is in motion with a velocity = v, the potential energy is at zero while the kinetic energy is at maximum.

Before this happen, the a body at rest (P.E = max) is set on motion, the kinetic energy gradually increases till it converts all the potential energy in the system to kinetic energy and then reverses back when the body goes to rest again.

In this case, before the batter hits the ball, the kinetic energy was at zero while the potential energy was at maximum. However, when he hits the ball and sets it into motion with a velocity V, the potential energy converts to kinetic energy and moves the ball with that energy till it has expanded it and comes to rest.

Potential Energy → Kinetic Energy → Potential Energy.

That's how the system keeps changing.

You might be interested in
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

h_{A} = 100 m

v_{A} = v_{A}

h_{B} = 150 m

v_{B} = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed v_{A}

final speed  v_{B} = 0

thus,

v_{B}² = v_{A}² - 2 g h

0 = v_{A}² - 2 g h

v_{A}² = 2 g h

v_{A} = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

8 0
2 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
2 years ago
An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the
Marina86 [1]

Answer:

2.88*10^{8} m/s

Explanation:

The speed of light is given by

c=\frac {2d}{t} and t=\frac {\theta}{\omega} hence

c=\omega\frac {2d}{\theta}

Speed of light is given by

c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s

4 0
2 years ago
A student bikes to school by traveling first dN = 1.00 miles north, then dW = 0.600 miles west, and finally dS = 0.200 miles sou
kompoz [17]
Refer to the diagram shown below.

Define unit vectors along the x and y axes as respectively \hat{i} \, and \, \hat{j}.

Then the three successive displacements, written in component form, are respectively
\vec{dN} = 1.0 \, \hat{j} \\&#10;\vec{dW} = -0.6 \, \hat{i} \\  \vec{dS} = -0.2 \, \hat{j}

The total displacement for the first leg of the trip is
\vec{d} = \vec{dN} + \vec{dW} + \vec{dS} \\ \vec{d}= 1.0\hat{j}-0.6\hat{i}-0.2\hat{j} \\ \vec{d}=-0.6\hat{i}+0.8\hat{j}

Answer:
\vec{d} = -0.6\hat{i}+0.8\hat{j}   or  (-0.6, 0.8)


6 0
2 years ago
You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper
enyata [817]

Answer:

5.76 round off to 6

Explanation:

wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

4 0
2 years ago
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