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GrogVix [38]
2 years ago
5

The term relative intensity is used to denote the amount of energy expended per minute.

Physics
1 answer:
sergeinik [125]2 years ago
7 0
I think the answer is True
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consider the free-body diagram. if you want the box to move, the force applied while dragging must be greater than the
NeTakaya

Answer:

Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

                                       F_{s}=\mu_{s}\eta

Where,

                             μ - coefficient of static friction

                             η - normal force acting on the body

When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.

So, in order to move the body, the applied force should be greater than the force of static friction.

6 0
3 years ago
A student in the Biomechanics class has decided that she would like to make her arms
Rufina [12.5K]

Answer:

what is heat and transfer

7 0
2 years ago
A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.
Romashka-Z-Leto [24]
<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + \frac{1}{2} a t²

In this question u = 0 , because car was at rest initially

Thus S =  \frac{1}{2} a t²

here S is displacement and a is the acceleration of car

Therefore  360 =  \frac{1}{2} a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = \frac{360x2}{3600}

or a = 0.2 m s⁻²

4 0
3 years ago
If you were trying to describe the difference between power and work you could say:
avanturin [10]
Power is the energy transferred or "WORK DONE" in one second
6 0
2 years ago
Read 2 more answers
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
2 years ago
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