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3 years ago

7

If I had a six pulley system and a resistance of 24,000 N, how much effort would I need to lift it?

1 answer:

lapo4ka [179]3 years ago

6 0

Suppose you could rig up a pair of pulley systems and get a friend, who is as strong as you, to pull one of them. You'd still have trouble, but you might be able to do it because each pulley would lift half the weight, or 1,000 pounds, and a quarter of that weight is 250 pounds

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How is work calculated when the force applied is not parallel to the displacement

Leno4ka [110]

Pretty sure it’s Force*Distance*Cos(theta)

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3 years ago

Read 2 more answers

Energy of position or location is called

Orlov [11]

The answer is B!
Explanation: Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.

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3 years ago

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

Mandarinka [93]

Answer:

The magnitude of minimum potential difference is 1800 V

Explanation:

Given:

Electric field $E = 3 \times 10^{6} \frac{V}{m}$

Gap between electrodes $d = 0.060 \times 10^{-2}$ m

For finding the minimum potential difference,

$\Delta V = E \times d$

$\Delta V = 3 \times 10^{6} \times 0.060 \times 10^{-2}$

$\Delta V = 1800$ V

Therefore, the magnitude of minimum potential difference is 1800 V

8 0

3 years ago

Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

kirza4 [7]

Answer:

The value is $U = 0.06 \ J$

Explanation:

From the question we are told that

The value of charge on each three point charge is

$q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C$

The length of the sides of the equilateral triangle is $r = 0.500 \$

Generally the total potential energy is mathematically represented as

$U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ]$

=> $U = k * 3 * \frac{q^2}{r}$

Here k is coulomb constant with value $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $U = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }$

$U = 0.06 \ J$

6 0

3 years ago

if 1.5kg block is pulled across a horizontal surface that has a coefficient of kintic friction of 0.60. if a block is pulled by

inysia [295]

Answer:

3.2N

Explanation:

Given parameters:

Mass of block = 1.5kg

Coefficient of kinetic friction = 0.6

Force of pull on block = 12N

Unknown:

Net force on the block = ?

Solution:

Frictional force is a force that opposes motion:

Net force = Force of pull - Frictional force

Frictional force = umg

u is coefficient of kinetic friction

m is the mass

g is the acceleration due to gravity

Frictional force = 0.6 x 1.5 x 9.8 = 8.8N

Net force = 12N - 8.8N = 3.2N

4 0

3 years ago