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Anarel [89]
3 years ago
12

In a lever, the effort arm is two times as long as the load arm. The resultant force will be

Physics
1 answer:
Taya2010 [7]3 years ago
5 0
So we want to know what the resultant force would be if on the lever the effort arm is as twice as long as the load arm. So in order for the lever to be in a state of balance it must be: F1*r1=F2*r2 where F1 is the effort arm and F2 is the load arm, r1 is the perpendicular distance from the force F1 to the pivot and r2 is the perpendicular distance from the force F2 to the pivot. For both sides of the equation to be equal, since r1=2*r2, F1*2*r2=F2*r2. Now we can divide by r2 and we get: 2F1=F2. When we divide by 2: F1=(1/2)*F2. So the force of the hand that is trying to lift the load is twice as small as the force of the load we are trying to lift. 
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Answer:

2.5N

Explanation:

since force is inversely proportional to square of the distance apart. that means the ratio of distance apart. so u have 12800×12800÷(6400××6400)=4/1

so the weight will be reduced to 1/4 of 10N

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Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D
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D. a telescope

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Which has the same meaning as air pressure?
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B) atmosphere pressure, i believe

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A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in
Nitella [24]

Answer:

Total number of lamps will be 4            

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to P=VI

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I=3.636A

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be n=\frac{15}{3.636}=4.125

As the lamp can not be in negative

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5 0
3 years ago
A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur
daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

4 0
3 years ago
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