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Mariulka [41]
2 years ago
8

In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s .The batter hits the ball, and

it leaves his bat with a speed of 24.0 m/s in exactly the opposite direction.
What is the magnitude of the impulse delivered by the bat to the ball?
If the bat is in contact with the ball for 1.4 ms , what is the magnitude of the average force exerted by the bat on the ball?
Physics
1 answer:
Deffense [45]2 years ago
8 0

Answer:

2.465Ns;1.76N

Explanation:

From Newton's Second Law:

There is a conservation of Momentum hence:

Momentum before the impact equals that after the impact.

Hence.

1. The Impulse is same as the change in Momentum which is;

0.145 ×17 =2.465Ns { note: 145g in kg is 0.145}

2. From Newton's third Law; force is the rate of change of impulse expressed mathematically as;

F = m×v / t

Where m×v is impulse and t is time given.

F = 2.465/ 1.4 = 1.76N

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The answer is as voltage increases current increases and therefore resistance would remain constant

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A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
Researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the fil
earnstyle [38]

Answer:

Minimum thickness; t = 9.75 x 10^(-8) m

Explanation:

We are given;

Wavelength of light;λ = 585 nm = 585 x 10^(-9)m

Refractive index of benzene;n = 1.5

Now, let's calculate the wavelength of the film;

Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene

Thus; λ_film = 585 x 10^(-9)/1.5

λ_film = 39 x 10^(-8) m

Now, to find the thickness, we'll use the formula;

2t = ½m(λ_film)

Where;

t is the thickness of the film

m is an integer which we will take as 1

Thus;

2t = ½ x 1 x 39 x 10^(-8)

2t = 19.5 x 10^(-8)

Divide both sides by 2 to give;

t = 9.75 x 10^(-8) m

8 0
2 years ago
When a carousel is in motion, the movement of the carousel horse can best be described as?
tatiyna
Supposing the carousel is rotating with constant speed, the movement is uniform angular motion.
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3 years ago
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
3 years ago
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