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pashok25 [27]
2 years ago
12

A pulley system used to lift car tires has a mechanical advantage of 11.2. If you pull on the pulley with a force of 150 N, how

much force does the pulley exert on the tire?
Physics
2 answers:
Novay_Z [31]2 years ago
7 0
1680 is the answer i got because i did this problem in apex
IgorC [24]2 years ago
5 0

Answer:

Load = 1680 N

Explanation:

Mechanical advantage of the pulley system, MA = 11.2

If a person pull on a pulley with a force of 150 N we have to find the amount of force the pulley exert on the tire. Mechanical advantage of a pulley system is given by the ratio of load to the effort force i.e.

IMA=\dfrac{load}{effort}

11.2=\dfrac{load}{150\ N}

Load = 1680 N

Hence, the force the pulley exert on the tire is 1680 Newtons.

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It would change the sign on the vector quantities and have no change to the scalar quantities
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6 0
2 years ago
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
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<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
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</span><span> E = -150 N/C (- because it is down and up is positive) 
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 a = -6,4*10^5 m/s^2 (downward) 
3 0
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When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p
Gnesinka [82]

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

U=mgh

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

6 0
2 years ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

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We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

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Large face measurements (D L)

       Length L = D= 2W

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The relationship is

    R₂ / R₁ = 2W²/L

6 0
2 years ago
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