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2 years ago

Find the efficiency of a machine that does 800 J of work if the input work is 2,000 J.

Physics

1 answer:

torisob [31]2 years ago

3 0

Answer:

Efficiency: 0.4 (40%)

Explanation:

The efficiency of a simple machine is given by the ratio

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{in}$ is the input work of the machine

$W_{out}$ is the output work of the machine

For the machine in this problem, we have:

$W_{out}=800 J$ is the output work

$W_{in}=2000 J$ is the input work

Therefore, the efficiency of the machine is

$\eta=\frac{800}{2000}=0.4$

Which can be also written as percentage:

$\eta=0.4\cdot 100 = 40\%$

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Answer:

A. The water decreases the friction between the floor and the feet

Explanation: Think about it like this, when your in the shower and water is on your skin, you can scrub it fluidly, but when the water dries in your towel, there is more friction when your rub your skin, this is because the molecules in water aren't as compact as solids, so anything acting against it, is most likely to disperse from it.

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2 years ago

A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

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Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

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2 years ago

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu

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A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. 

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N

Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2

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2 years ago

Read 2 more answers

Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who

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Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time