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3 years ago

How long does it take a 22 kW steam engine to do 5.6 × 107 J of work? Answer in units of s.

Physics

1 answer:

Lena [83]3 years ago

5 0

<h2>Time needed is 2545.45 seconds.</h2>

Explanation:

We know equation for power

$\texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}$

Here we need to find time when for a 22 kW steam engine to do 5.6 × 10⁷ J of work.

Work = 5.6 × 10⁷ J

Power = 22 kW = 22 x 10³ W

Substituting

$\texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}\\\\22\times 10^3=\frac{5.6\times 10^7}{\texttt{Time}}\\\\\texttt{Time}=2545.45s$

Time needed is 2545.45 seconds.

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A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?

notsponge [240]

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

4 0

2 years ago

Can anyone help me with this please

Anastasy [175]

Answer:

1270 J

Explanation:

Recall that the mechanical energy of a system is the addition of the Potential energy and the Kinetic energy at any given time.

As the skier descends, potential energy is converted into kinetic energy, but the total mechanical energy should remain the same.

We see that it is not the case, so that difference is what has gone into thermal energy; 19500 J - 18230 J = 1270 J

7 0

3 years ago

A brick is lying on a table in a state of static equilibrium. If the mass of the brick is 7.52 kilograms, what is the normal for

WARRIOR [948]

Answer:

-mg = -7.52*9.8 = - 73.696N

7 0

2 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

A) Two workers are trying to move a heavy crate. One pushes onthe crate with a force A, which has amagnitude of 445 newtons and

Reptile [31]

Answer:

Divide then multiply or multiply then divide

Explanation:

to get the answer of a and b

3 0

2 years ago