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snow_tiger [21]
3 years ago
15

-A: Find the force that must be exerted on the rod to maintain a constant current of 0.156 A in the resistor? (In mN)

Physics
1 answer:
Ksenya-84 [330]3 years ago
4 0

A) Force: 0.0528 N

B) Power dissipated: 0.307 W

C) Power delivered: 0.307 W

Explanation:

A)

The  force experienced by a current-carrying wire in a magnetic field is given by

F=ILB

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

In this problem, we have:

L = 0.451 m

B = 0.751 T

I = 0.156 A

Therefore, the force is

F=(0.156)(0.451)(0.751)=0.0528 N

B)

The rate of energy dissipation in the resistor is the power dissipated in the resistor, and it is given by

P=I^2R

where

I is the current in the resistor

R is the resistance

For the wire in this problem,

I = 0.156 A

R=12.6\Omega

Therefore, the power dissipated is

P=(0.156)^2(12.6)=0.307 W

C)

The mechanical power delivered to the rod is given by

P=VI

where

V is the potential difference across the rod

I is the current in the rod

The potential difference across the rod must be equal to the potential difference across the resistance, which can be  found by using Ohm's law:

V=RI=(12.6 \Omega)(0.156 A)=1.97 V

Therefore, the power delivered to the rod is

P=(1.97 V)(0.156 A)=0.307 W

This power is equal to the power dissipated on the resistor: this is due to the law of conservation of energy, in fact the total energy must remain constant, so here the electric energy is transformed into mechanical energy of motion of the rod.

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What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a
Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2

v = r ω

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}

I = \dfrac{m(2gh - v^2)r^2}{v^2}

I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}

I =mr^2(0.287)

I = 0.287 MR²

3 0
3 years ago
A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?
PSYCHO15rus [73]
5x2.2/3 Because that's how you calculate velocity.

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3 years ago
What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.
djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec    

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity g=9.8m/sec^2

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that v^2=u^2+2gh

v^2=0^2+2\times 9.8\times 150

v=54.22m/sec

So the velocity of the ball when its hit the ground will be 54.22 m/sec

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3 years ago
1. seesaw
Sergeeva-Olga [200]

Answer:

1  seesaw = first class lever

2. pencil and sharpener = wheel and axle

3. bottle opener = second class lever

4. forearm = third class lever

5. nail = wedge

Explanation:

The given objects are classic examples of simple machines.

Lever: There are 3 types of levers and they depend on where the load, applied force, and fulcrum are.

First class levers have the fulcrum in between the load and the applied force. In other words, the load and applied force are at opposite ends. The seesaw is a good example of this. Other examples would be, pliers, scissors, and the like.

Second class levers have the fulcrum and the applied force at the opposite ends. So in this case, the load is found in between. Examples of this would be a bottle opener or a crowbar.

For third class levers, the load and the fulcrum are at opposite ends and the applied force is in between. The forearm is a good example, so is a stapler, or even a broom.

A wedge is another simple machine. It is thick at one end and it gets thinner towards the other end, or it usually has a sharp end. Other examples of this would be an axe, or a knife.

A sharpener is actually a compound machine. The old-fashioned type of pencil sharpener, the one you crank makes use of a wheel and axle and a wedge. A wheel and axle is usually a machine that makes use of two circular parts; a wheel and a rod that is attached to its center.

4 0
3 years ago
A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m
Keith_Richards [23]

Answer:

42.69 N and 18.07 N

Explanation:

We are given that

Mass of ladder=6.2 kg

Length of ladder=1.97 m

Distance of Sawhorse A from one end=0.64 m

Distance of sawhorse B from other end=0.17 m

Let center of Ladder=\frac{1.97}{2}=0.985 m

Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m

Distance of sawhorse B from center of ladder=0.985-0.17=0.815  m

Force one ladder due to gravity=mg=6.2\times 9.8=60.76N

Where g=9.8 m/s^2

Torque applied on Sawhorse A=0.345F_a

Torque applied on Sawhorse B=0.815F_b

In equilibrium

0.345F_a=0.815F_b

F_b=\frac{0.345}{0.815}F_a

Total force=F_a+F_b

F_a+\frac{0.345}{0.815}F_a=60.76

\frac{0.815F_a+0.345F_a}{0.815}=60.76

\frac{1.16}{0.815}F_a=60.76

F_a=\frac{60.76\times 0.815}{1.16}=42.69 N

F_b=\frac{0.345}{0.815}\times 42.69=18.07 N

8 0
3 years ago
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