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hodyreva [135]
3 years ago
15

10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends

soccer ball towards the goal.
G
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S
S
Formula

11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?

G
U
E
S
S
Formula

please I need help I don't understand it and I had to deliver it yesterday helpp:(

Physics
1 answer:
maxonik [38]3 years ago
5 0

Answer:

10. 65 N

11. 35 N

Explanation:

10. Determination the force that sends the soccer ball towards the goal.

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Initial velocity (u) of the ball = 0 m/s

Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s

Acceleration (a) of the ball =?

a = (v – u) /t

a = (32.5 – 0) / 0.21

a = 32.5 / 0.21

a = 154.76 m/s²

Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:

Mass (m) of the ball = 0.42 kg

Acceleration (a) of the ball = 154.76 m/s²

Force (F) =?

F = ma

F = 0.42 × 154.76

F = 65 N

Thus, the force that sends soccer ball towards the goal is 65 N

11. Determination of the force exerted by the rocket on the satellite.

We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:

Initial velocity (u) of satellite = 0 m/s

Final velocity (v) of satellite = 0.63 m/s

Time (t) = 1296 s

Acceleration (a) of the satellite =?

a = (v – u) /t

a = (0.63 – 0) / 1296

a = 0.63 / 1296

a = 4.861×10¯⁴ m/s²

Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:

Mass (m) of the satellite = 72000 Kg

Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²

Force (F) =?

F = ma

F = 72000 × 4.861×10¯⁴

F = 35 N

Thus, the force exerted by the rocket on the satellite is 35 N

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3. A large passenger ship required a force of 1,600,000 N to move 2000 m. How much work is done on the ship?​
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3,200,000,000 J

Explanation:

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Here, we have F = 1,600,000 N and s = 2000 m, so our work will be

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A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component
agasfer [191]

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

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as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

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           m_B  λ_B = m_A  λ_A

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8 0
3 years ago
A 0.026 kg bullet is fired straight up at a falling wooden block that has a mass of 5.0 kg. The bullet has a speed of 750 m/s wh
Law Incorporation [45]

Answer:

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a = acceleration = - g

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Using the kinematics equation

v_{f} = v_{o} + at \\v_{f} = 0 + (-g)t\\v_{f} = - gt

m = mass of the bullet = 0.026 kg

v_{b} = velocity of block just before collision = 750 m/s

M = mass of the block = 5 kg

V = final velocity of bullet block after collision = gt

Using conservation of momentum

m v_{b} + Mv_{f} = (m + M) V\\(0.026) (750) + (5) (- gt) = (0.026 + 5) (gt)\\19.5 - 49 t = 49.2548 t\\t = 0.198 s

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