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4 years ago

Carlotta does 2000 j of work on a machine. The machine does 500 J of work what is the efficiency of this machine

Physics

1 answer:

almond37 [142]4 years ago

8 0

Efficiency of a machine = (useful output energy) / (total input energy)

Efficiency = (500 J output work) / (Carlotta's 2,000 J input work)

Efficiency = (500 J) / (2,000 J)

Efficiency = 0.25

Efficiency = 25%

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A skater goes into a spin with her arms pulled in (close to her body). When she stretches her arms out: A) her angular speed ω d

musickatia [10]

Answer:

When she stretches her arms out, B) her angular speed ω increases due to her moment of inertia decreasing

Explanation:

The angular momentum of a rotating object is defined as the product of its moment of inertia and angular speed.

L = I ω

where

L is the angular momentum
I is the moment of inertia
ω is the angular speed

According to the principle of conservation of angular momentum, if there is no external torque, angular momentum of the skater must remain conserved. If the initial and final moment of inertia is I_i and I_f while corresponding angular velocities are ω_i and ω_f , then the principle of conservation of angular momentum can be expressed as the following equation:

(I_f) (ω_f) = (I_i) (ω_i)

ω_f / ω_i = I_i / I_f

From the expression above, we can see that if the moment of inertia decreases, angular velocity would increase to conserve angular momentum of the skater.

Therefore, When she stretches her arms out, her angular speed ω increases due to her moment of inertia decreasing.

5 0

4 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

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3 years ago

Different between pulling and pushing force

yuradex [85]

Answer:

Push and pull are both forces, but the difference is in their direction at which it is applied. If the force applied in the direction of motion of the particle then we call it push. If that force applied in the direction OPPOSITE to the motion of the particle then it it termed as pull

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3 years ago

A child is sliding on a sled at 1.3 m/s to the right. You stop the sled by pushing on it for 0.80 s in a direction opposite to i

Shalnov [3]

Answer:

find acceleration first

a = vf - vi / t

a = 0 - 1.5 / 0.5s (vf is zero)

a = -3 (negative sign indicates that acceleration is decreasing)

so F=ma

F = 35 x -3

F = -105 N (here negative sign indicates that u have to apply a force opposite to the boy's direction i.e from the left or towards the right)

Explanation:

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4 years ago

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AnnyKZ [126]

Answer:

Explanation: what is the question?

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3 years ago