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3 years ago

14

The electric motor has an input energy of 50,000 joules each second. The motor transfers 35,000 joules of useful energy each sec

ond. Calculate the efficiency of the electric motor (show workig out too)

1 answer:

Sergio039 [100]3 years ago

6 0

W(out)/w(in)
35,000J/50,00J=.7
.7 *100%=70 % efficient

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This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the

valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2$

Equation of motion

$v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s$

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

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3 years ago

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers

jek_recluse [69]

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v = velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

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2 years ago

Which formula can be used to solve problems related to the first law of thermodynamics?

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Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

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