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katen-ka-za [31]
3 years ago
5

I need a short written description of Newton's three laws.

Physics
1 answer:
Tom [10]3 years ago
6 0
Newton’s first law of motion, also called the law on inertia, states that an object continues in its state of rest or of uniform motion unless compelled to change that state by an external force.Newton’s second law of motion states that if a net force acts on an object, it will cause an acceleration of that object.Newton’s third law of motion<span> states that for every action there is an equal and opposite reaction. hope this wasnt two long!</span> 
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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas
Nookie1986 [14]

The maximum amount of work performed is

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Explanation:

The efficiency of a real heat engine is given by the equation:

\eta = 1-\frac{T_C}{T_H} (1)

where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

where

Q_Cis the heat released to the cold reservoir

So the previous equation can be also written as

\eta=\frac{W_{max}}{W_{max}+Q_C} (2)

By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Learn more about work and heat:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

8 0
3 years ago
How does an increase in temperature generally affect the rate of a reaction?
Zigmanuir [339]

Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.

If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.

7 0
3 years ago
when the mass of an object is increased it would decrease its acceleration in the force is left alone
mojhsa [17]

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

a=\frac{F}{m}

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

4 0
3 years ago
A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the
Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,                                

length of rod = 10 m                

height of drop = 4.89 m          

Earth’s magnetic field =  12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam                      

v = \sqrt{2gh}

v = \sqrt{2\times 9.8 \times 4.89}

v = 9.79 m/s                        

emf of the beam

e = B l v                              

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0
3 years ago
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