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3 years ago

Consider this situation: Four ropes, each attached to the end

Physics

1 answer:

faust18 [17]3 years ago

5 0

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is $mg$ , where m is the mass of the object and g is the acceleration of gravity.
Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

Learn more about forces:

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A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located

zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

$\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241 \times 10^{-19 } \ C$

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

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2 years ago

A ball is spun around in circular motion such that it completes 50 rotations in 25 s.

Leokris [45]

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

$T=\dfrac{25}{50}\\\\T=0.5\ s$

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

$f=\dfrac{50}{25}\\\\f=2\ Hz$

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3 years ago

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The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie

pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

$\bf \longrightarrow {v}^{2} = {u}^{2} + 2as \\ \\ \rm \longrightarrow {0}^{2} = {15}^{2} + 2 \times a \times 63 \\ \\ \rm \longrightarrow 0 = 225 + 126a \\ \\ \rm \longrightarrow 126a = - 225 \\ \\ \rm \longrightarrow a = - \dfrac{225}{126} \\ \\ \rm \longrightarrow a = - 1.78 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

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2 years ago