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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 do echolocation and ultrasounds have in common?

Natali5045456 [20]

They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.

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2 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

2 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

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3 years ago

The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

Setler [38]

Answer:

Explanation:

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3 years ago

Select the correct answer.

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

We’ll see the bell move, but we won’t hear it ring.

We won’t see the bell move, but we’ll hear it ring.

We’ll see the bell move and hear it ring.

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Explanation:

so, the answer to the question is

We'll see the bell move, but we won’t hear it ring.

5 0

2 years ago

Read 2 more answers