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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.
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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
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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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4 0
4 years ago
A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest orde
True [87]

Answer:

The order of maximum is   n = 5

Explanation:

From the question we are told that

  The  diffraction grating is  k  =  300 lines per mm  =  300000 lines per m

   The wavelength is  \lambda  =  630 \  nm  =  630 *10^{-9} \  m

   Generally the condition for constructive interference is mathematically represented as

      dsin \theta = n  * \lambda

Here n is the order maximum

d is the distance the grating which is mathematically represented as

    d =  \frac{1}{k}

=>   d =  \frac{1}{300000}

=>    d =  3.3*10^{-6}\  m

So

   n  = \frac{dsin \theta}{ \lambda}

at maximum  sin\theta  =  1

     n  = \frac{d}{\lambda}

=>   n  = \frac{3.3*10^{-6}}{630 *10^{-9}}

=>   n = 5

 

8 0
3 years ago
A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an
anyanavicka [17]

Answer:

The value of tangential acceleration \alpha_{t} =  40 \frac{m}{s^{2} }

The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

Explanation:

Angular acceleration = 50 \frac{rad}{s^{2} }

Radius of the disk = 0.8 m

Angular velocity = 10 \frac{rad}{s}

We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

\alpha = angular acceleration

⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

Radial acceleration is given by

\alpha_{r} = \frac{V^{2} }{r}

Where V = velocity of the disk = r \omega

⇒ V = 0.8 × 10

⇒ V = 8 \frac{m}{s}

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\alpha_{r} = \frac{8^{2} }{0.8}

\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

7 0
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The speed of x-rays is A.faster than light B.slower than the speed of gamma rays C.Same as the speed of radio waves D.same as th
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5 0
1 year ago
La cantidad de materia en un objeto.
Scrat [10]

La respuesta correcta es Masa

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