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3 years ago

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

Physics

2 answers:

Nana76 [90]3 years ago

5 0

Answer:

Efficiency: 0.4 (40%)

Explanation:

The efficiency of a simple machine is given by the ratio

n = w¹/w²

where

w² is the input work of the machine

w¹ is the output work of the machine

For the machine in this problem, we have:

w¹ = 800j is the output work

w² = 2000j is the input work

Therefore, the efficiency of the machine is

n = 800/2000 = 0.4

Which can be also written as percentage:

n = 0.4 × 100 = 40%

hope it helps you

IgorLugansk [536]3 years ago

3 0

Answer:

Efficiency: 0.4 (40%)

Hope it helps you

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A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago

If 0.5 A is flowing through a household light

stiks02 [169]

Answer:

60W

Explanation:

P=IV=0.5x120

P =60W

5 0

3 years ago

A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

Στ = 0

τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

+ W d₁ - w_cat d₂ = 0

d₂ = W / w d₁

d₂ = M /m d₁

d₂ = 5.00 /2.9 0.850

d₂ = 1.466 m

6 0

4 years ago

You are driving your car, and the traffic light ahead turns red. you apply the brakes for 2.26 s, and the velocity of the car de

Lorico [155]

Let the initial velocity of the car be u.

Final velocity of the car (v) = 5.43 m/s

deceleration (a) = - 2.78 m/s^2

Time taken (t) = 2.26 s

Using the first equation of motion:

v = u + at

u = v - at

$u = 5.43 - (-2.78 \times 2.26)$

u = 5.43 + 6.28

u = 11.71 m/s

Let the car's displacement be x.

Using second equation of motion:

$x = ut + \frac{1}{2}at^2$

$x = 11.71 \times 2.26 - \frac{1}{2} \times 2.78 \times 2.26^2$

x = 26.4646 - 7.0995

x = 19.3651 meters

Hence, the displacement of the car is 19.36 meters

6 0

3 years ago

A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 45.0 deg

Romashka [77]

Answer:

a. 139.748J

b.3.486m/s

c.zero

Explanation:

a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°

Potential energy during release is calculated as: $PE=mgh$

#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.

Find the height when the child is pulled back:

$cos 45\textdegree=y/2.10\\\\y=1.48m$

#therefore,vertical difference is 2.1m-1.48m=0.62m

$\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J$

#Hence the potential energy during release is 139.748J

b. From a, above, we have PE=139.748J, M=23.0kg.

At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:

$KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s$

#Hence the velocity at the bottom of the swing is 3.486m/s

c. Work is calculated as the product of force by distance.

From a, b above we have mass as 23.0kg .

-since the distance of the ropes remained constant the change in distance is zero:

$W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0$

Therefore the work in the ropes is 0,zero.

8 0

3 years ago