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guajiro [1.7K]
3 years ago
15

A machine has an efficiency of 70%. How much work dose the machine do when 20,000 J of work is done on it ?

Physics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer: The machine does 14.000 J

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How fast must the space shuttle go to cover 20,000 meters in 4.0 seconds?
Advocard [28]

Answer:

5000 m/s

Explanation:

7 0
3 years ago
A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest
Ymorist [56]

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

3 0
2 years ago
A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what
andreyandreev [35.5K]

Answer:

The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.

when angle is zero then flux is maximium because when angle zerocos is maximium

8 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.
sineoko [7]

Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

5 0
3 years ago
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