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3 years ago

In a simple machine, the energy input is 120 J. If the efficiency of the machine is 80%, calculate the energy output

Physics

1 answer:

pashok25 [27]3 years ago

4 0

Answer:

<h2>96 Joules</h2>

Explanation:

We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.

Solution,

Energy input of system = 120 J

Efficiency = 80% = $\frac{80}{100} = 0.8$

Now,

According to definition,

Efficiency = $\frac{output}{input}$

Cross multiplication:

$output \: = \: 0.8 \times 120$

Calculate the product

$output \: = 96 \: joules$

Hope this helps...

Good luck on your assignment...

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What is the order of magnitude of the distance of Sun to nearest star in meters?

neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

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Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

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4 0

2 years ago

Read 2 more answers

a cart has 30J or potential energy at the top of a 20m high hill. Assuming no energy is transformed to another form of energy, h

topjm [15]

Answer:

i would say 20J

Explanation:

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3 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

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2 years ago

A 1060-kg car moving west at 16 m/s collides with and locks onto a 1830-kg stationary car. determine the velocity of the cars ju

iren [92.7K]

M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s

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3 years ago

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660 N

==>T× 0.7660 = 29.4

==> T= 38.38 N

8 0

3 years ago