All categories

3 years ago

A steam engine does 12 kJ or work in 55 s. how much power does the engine produce?

Physics

2 answers:

allsm [11]3 years ago

8 0

Answer:

Power, P = 218.18 watts

Explanation:

It is given that,

Work done by the steam engine, $W=12\ kJ=12\times 10^3\ J$

Time taken to do work, t = 55 seconds

Power produced by the engine is given by work done divided by time taken i.e.

$P=\dfrac{W}{t}$

$P=\dfrac{12\times 10^3\ J}{55\ s}$

P = 218.18 watts

Hence, this is the required solution.

Georgia [21]3 years ago

6 0

P=w/t
12kj=12000J
the answer is p=12000/55=218.18

You might be interested in

An object is placed at zero zero on a Number line. It moves three units to the right, then four units to the left, and then 60 u

STatiana [176]

Answer:

the displacement of the object is 5 units

Explanation:

The computation of the displacement of the object is shown below:

= Move to the right + move to the right - move to the left

= 6 units + 3 units - 4 units

= 9 units - 4 units

= 5 units

Hence, the displacement of the object is 5 units

7 0

3 years ago

William Ferrel:

balu736 [363]

Answer:

William Ferrel created a tide-prediction machine.

Explanation:

William Ferrel create a machine in late 19th century that was the best combination of mechanical parts and computer coding.
It was a mechanical analog computer that could predict the ebb of tides and even the height of tides that could be irregular.
It was widely used for marine networks and navigation. Later on many improvisations and additional features were added on it.
During the world war times, this tide prediction machine was of great use for military purpose.

3 0

3 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

3 years ago

On Earth, a kangaroo jumping will eventually return to ground due to the unbalanced force of gravity. What law does this illustr

Kryger [21]

Newton’s law is the answer

6 0

3 years ago

Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$

Explanation:

Looking at the diagram uploaded we see that there are two forces acting along the x-axis on the fixed support

These force are 400 N and $A_{(x)}$ [ i.e the reactive force of 400 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(x)} - 400 = 0$

=> $A_{(x)} = 400 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the y-axis on the fixed support

These force are 500 N and $A_{(y)}$ [ i.e the force acting along the same direction with 500 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(y)} + 500 = 0$

=> $A_{(y)} = -500 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the z-axis on the fixed support

These force are 600 N and $A_{(z)}$ [ i.e the reactive force of 600 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(z)} - 600 = 0$

=> $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

$\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=> $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

$\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=> $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

$\sum M_z = M_z = 0$

=> $M_z = 0 \ N\cdot m$

8 0

3 years ago