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madam [21]
3 years ago
13

Which mathematical formula would you use to calculate the IMA of a wheel?

Physics
1 answer:
Crazy boy [7]3 years ago
3 0
As shown in the figure, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle. Any crank-operated device is an example of a wheel and axle. Force applied to a wheel exerts a force on its axle.
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To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newt
Firdavs [7]

Answer: A.E.

Explanation: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton's law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word "weight" often replaces "mass," as in "My weight is seventy-five kilograms"

6 0
3 years ago
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A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
3 years ago
Earth travels fastest in January and slowest in July. What is the most likely explanation for this?
Keith_Richards [23]

Answer:

Earth is nearest the Sun in July and farthest away in July.

Explanation:

3 0
3 years ago
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To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
3 years ago
The current supply from generating plants is alternating current. True or false?
Leno4ka [110]
The answer is True since thats what usually goes on
3 0
3 years ago
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