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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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On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

3 years ago

Read 2 more answers

Is 40,000 °C hotter or colder than 3,600 °C????

Amiraneli [1.4K]

Yes because if 0*c equals 32*F than the higher the number the hotter it is

4 0

3 years ago

Read 2 more answers

A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

Acceleration = (-28 m/s) / (13 sec)

Acceleration = -2.15 m/s²

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3 years ago

Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical tran

horsena [70]

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

I₁ = I₀ / 2

I₁ = 845/2

I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

Cos² θ = I / I₁

Cos θ = √ I / I₁

Cos θ = √ (275 / 422.5)

Cos θ = 0.80678

θ = cos⁻¹ 0.80678

θ = 36.2º

This is the angle between the two polarizers

8 0

3 years ago

A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

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3 years ago