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2 years ago

How can you make the work output of machine greater than the work input

Physics

1 answer:

ivolga24 [154]2 years ago

5 0

Answer:

The only way to increase the work output is to increase the work input. You can't get more work out of a machine than you put into it. The moving parts of a machine must use some of the work input to overcome friction. Because of friction, the work output of a machine is always less than the work input.

Explanation:

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Convection occurs when thermal energy is transferred by the movement of

madam [21]

Answer:

Convection is the transfer of thermal energy from one place to another by the movement of gas or liquid particles. How does this happen? As a gas or liquid is heated, the substance expands. This is because the particles in liquids and gases gain kinetic energy when they are heated and start to move faster.

6 0

3 years ago

What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (see section 17.6;

antiseptic1488 [7]

<h3><u>Answer;</u></h3>

volume = 6.3 × 10^-2 L

<h3><u>Explanation</u>;</h3>

Volume = mass/density

Mass = 0.0565 Kg,

Density = 900 kg/m³

= 0.0565 kg/ 900 kg /m³

= 6.3 × 10^-5 M³

but; 1000 L = 1 m³

Hence, <u>volume = 6.3 × 10^-2 L</u>

8 0

3 years ago

A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ign

Gemiola [76]

B- the acceleration is greater for the more massive rock

5 0

2 years ago

Read 2 more answers

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

3 years ago

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and

Paladinen [302]

Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc
or
$r = \frac{m v^{2}}{F}$

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton

To find = radius of track=?
$r = \frac{m v^{2} }{F}$
$r = \frac{1600*13*13}{3300}$
r = 81.94 m
Therefore, radius of track is 81.94 m

4 0

3 years ago

Read 2 more answers