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ololo11 [35]
3 years ago
10

The force that resists moving one object against another is _____. friction push energy work

Physics
2 answers:
masha68 [24]3 years ago
8 0
Your answer would be friction. 
Novosadov [1.4K]3 years ago
6 0
The answer is A. Friction.

Friction is the resisting force between two objects that move against one another.

Hope this helps! :)


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An engineer in India (standard household voltage = 220 volts) is designing a transformer for use on her
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He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns

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3 years ago
A light bulb is connected to a 2V supply and experiences a current of 6.4A. What is the power rating of the bulb?
Olenka [21]

Answer:

12.8 Watts

Explanation:

P = VI

P = (2 V) (6.4 A)

P = 12.8 W

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4 years ago
Why do people have symptoms in their bodys
weeeeeb [17]

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For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 78kg adult receive?
kari74 [83]
RDA stands for Recommended Daily Allowance. To determine the amount needed of a certain adult per day, we simply multiply the mass of the adult to the value of RDA. For this case, we do as follows:

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6 0
4 years ago
A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F
EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B =  1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ)   eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0
3 years ago
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