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balu736 [363]
3 years ago
5

A pitcher throws a softball toward home plate. When the ball hits the catcher’s mitt, its horizontal velocity is 32 meters/sec

ond. The softball’s velocity goes to 0 meters/second in 0. 8 seconds when caught. If the softball has a mass of 0. 2 kilograms, what’s the force of the impact? Use F = ma, where. The force of the impact is newtons.
Physics
1 answer:
vladimir1956 [14]3 years ago
7 0

The force of impact of the softball at the given conditions is 8 N.

The given parameters:

  • <em>Horizontal velocity of the ball, v = 32 m/s</em>
  • <em>Change in time of motion of the ball, Δt = 0.8 s</em>
  • <em>Mass of the ball, m = 0.2 kg</em>

The force of impact of the softball is calculated by applying Newton's second law of motion as follows;

F = ma\\\\F = m \times \frac{\Delta v}{\Delta t} \\\\F = 0.2 \times \frac{32 - 0}{0.8 - 0} \\\\F = \frac{0.2 \times 32}{0.8} \\\\F = 8 \ N

Thus, the force of impact of the softball at the given conditions is 8 N.

Learn more about Newton's second law of motion here: brainly.com/question/25307325

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Answer:

B

Explanation:

momentum = mass * velocity

velocity = initial speed + acceleration * time where initial speed =0, acceleration = 9.81, time = 4

8 0
3 years ago
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Which of the following statements about flux are true?
vekshin1
All of these statements about flux are true, except for the last one - <span>When turning a surface inside of an electric field, the maximum flux is achieved if the electric field vector and the surface vector are perpendicular. 
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8 0
3 years ago
In a classical carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast en
g100num [7]

Answer:

- the speed of a person "stuck" to the wall is 14.8 m/s

- the normal force of the wall on a rider of m=54kg is 1851 N

- the minimum coefficient of friction needed between the wall and the person is 0.29

Explanation:

Given information:

the radius of the cylindrical room, R = 6.4 m

the room spin with frequency, ω =  22.1 rev/minutes = 22.1 \frac{2\pi }{60} = 2.31 rad/s

mass of rider, m = 54 kg

the speed of a person "stuck" to the wall

v = ω R

  = 2.31 x 6.4

  = 14.8 m/s

the normal force of the wall on a rider

F = m a

a  = ω^2 R

   =  \frac{v^{2} }{R^{2} } R

   = \frac{v^{2} }{R}

F = \frac{mv^{2} }{R}

  = \frac{(54)(14.8)^{2} }{6.4}

  = 1851 N

the minimum coefficient of friction needed between the wall and the person

F(friction) = μ N

W =  μ N

m g =  μ \frac{mv^{2} }{R}

g = μ \frac{v^{2} }{R}

μ = \frac{gR}{v^{2} }

  = \frac{(9.8) (6.4)}{14.8^{2} }

  = 0.29

5 0
3 years ago
Air breaks down and conducts charge as a spark if the electric field magnitude exceeds V/m. Determine the maximum charge that ca
Ymorist [56]

Answer:

Explanation:

Suppose

Magnitude of Electric Field is E V/m

Area of the cross-section is A

capacitor C=\frac{\epsilon A}{d}

Distance between Area of capacitor is d

Maximum Charge stored is

Q_{max}=capacitor\times Potential\ Difference

Potential\ Difference=electric\ Field\times distance=E\times d

Q_{max}=C\times Ed=CEd

Q_{max}=\frac{\epsilon _0A}{d}\times Ed

Q_{max}=\epsilon _0AE

 

3 0
4 years ago
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2

For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2

The second distance, r₂, can be determined from sound intensity formula given as;

I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 =  \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 =   \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m

Therefore, the second distance of the sound from the source is 431.78 m.

7 0
3 years ago
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