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Lemur [1.5K]
3 years ago
15

A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?

Physics
1 answer:
Mkey [24]3 years ago
5 0

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).  

On Earth, that's (1600 kg) x (9.8 m/s²)  =  15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

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3 years ago
Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward
Dafna11 [192]

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0.54

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Draw a free body diagram.  There are 5 forces on the desk:

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Friction force Fnμ pushing left

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If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in
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Answer:

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In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

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3 years ago
A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r
Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr 
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total charge = 4π * ρ0 (r³/3 + r^4/(4R)) 
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total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3 
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B) E = kQ/d² 
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C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr 
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which could be expressed other ways. 

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that 
r = R for a min/max (and we know it's a max since r = 0 is a min). 

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

4 0
3 years ago
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