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3 years ago

Why are some machines less than 100% efficient?

Physics

2 answers:

evablogger [386]3 years ago

7 0

Answer:

The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent. The closer to 100 percent a machine's efficiency is, the better it is at reducing friction

Explanation:

Contact [7]3 years ago

3 0

Answer:

Not all machines work effectevly.

Explanation:

A disrepare objest may not work as vigorus as a new object.

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What are the 6 uses of electromagnet?

Serhud [2]

I would say x-rays, microwave radiation, radio waves, Ultraviolet radiation, and Gamma rays.

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3 years ago

Read 2 more answers

what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

Evgen [1.6K]

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

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1 year ago

What is the maximum speed at which a car can safely travel around a circular track of radius 55.0 m If the coefficient of fricti

Stella [2.4K]

Answer:

The maximum speed of the car should be 13.7 m/s

Explanation:

For the car to travel at a maximum safe speed , the frictional force acting should be maximum and at the same time should provide the necessary centripetal force.

Let 'k' (=0.3502) be the coefficient of friction and 'N' be the normal force acting on the surface.

Then ,

N = mg , where 'm' is the mass of the body and 'g'(=9.8) is the acceleration due to gravity.

∴ Maximum frictional force , f = kN = kmg

Centripetal force that should act on the car to move with maximum possible speed is -

$F = \frac{mv^{2} }{r}$ , where 'v' is the velocity of the car and 'r'(=55m) is the radius of circular path.

Equating the 2 forces , we get -

$\frac{mv^{2} }{r} = kmg$

∴ $v = \sqrt{krg}$

Substituting all the values , we get -

v = 13.7 m/s.

5 0

3 years ago

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

guajiro [1.7K]

The correct answer to the question is- $2.2\ \mu C$

CALCULATION:

As per the question, the electric field generated by the source charge is 1236 N/C at a distance of 4 m.

Hence , electric field E = 1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

Electric field E = $\frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}$

Here, $\epsilon_{0}$ is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

Q = $E\times 4\pi \epsilon_{0}\times R^2$

= $1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb$

= $2197.33\times 10^{-9}\ C$

= $2.19733\times 10^{-6}\ C$

= $2.2\ \mu C$

Hence, the charge of source is $2.2\ \mu C$

3 0

3 years ago

Read 2 more answers

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

4 0

3 years ago