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tino4ka555 [31]
3 years ago
6

SECTION BTHEORY QUESTIONS(1a) Define the following(a) Work:​

Physics
1 answer:
Vesna [10]3 years ago
6 0

Work is the amount of energy transferred

Explanation:

In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.

For instance, work is done when a force is applied on an object. The work done on the object is given by:

W=Fd cos \theta

where

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

We notice the following:

  • No work is done when the force is perpendicular to the displacement (cos 90^{\circ}=0)
  • The work is maximum when the force is parallel to the displacement

Whenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel
Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

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3 years ago
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What is the mission of the mars Land Rover, and why is it important?
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3 years ago
A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula
lisov135 [29]

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

F_{N} = ma = F_{applied} - f_{fr}

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, m=15\ kg , F_{applied} =45\ N and f_{fr} = 30\ N

Substituting these values in above formula, we get

15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2.

Thus, the acceleration of an object is 1\ m/s^2.


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