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vlabodo [156]
3 years ago
10

According to Newton's second law, the acceleration of an object depends on its_ and the amount of _ applied to it.​

Physics
1 answer:
o-na [289]3 years ago
8 0
<h3>Solution:</h3>

Firstly, we should find out what is Newton's Second Law of Motion.

It states "The rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force."

In other words, the rate of change of momentum of an object is the acceleration of that object.

According to Newton's Second Law of Motion, F = ma, where F is the force, m is the mass and a is the acceleration.

So, F ∝ a

Also, F = ma

or, F/m = a

or, 1/m ∝ a.

So, we see that from Newton's Second Law, the acceleration of an object is directly proportional to the force acting upon the object, and inversely proportional to the mass of the object.

So, the acceleration depends on two variables:

  1. Mass of the object
  2. Amount of force applied to it
<h3>Answer:</h3>

According to Newton's second law, the acceleration of an object depends on its <u>mass</u> and the amount of <u>force</u> applied to it.

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A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Scilla [17]
Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in  N ]

From given above, area of cross section = &pi; r^2 = 100 &pi; =314 mm^2
(i) Stress,
&sigma;
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
&epsilon;
= ratio of extension / original length
= &sigma; / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= &epsilon; * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
4 years ago
If 270 wats of power is used in 42 seconds how much work is done ?
pentagon [3]

Answer:

power is given by work done divide time taken

P=w/t

therefore Pt=w

270x42=11,340.J

6 0
4 years ago
I need Help ASAP, and actually help, not just for the points.
UNO [17]

Answer:

750 force

Explanation:

I have never worked with force but I can guess using the formula. 1500, which is the mass, multiplied by the acceleration, 0.5, would equal 750 force, if being applied by the equation listed, Force= mass×acceleration

7 0
3 years ago
Read 2 more answers
A ball falls for 9.0 s increasing its kinetic energy by 2700. If the force acting on the ball in 6.0 N find the following quanti
Ivanshal [37]

Answer:

Change in velocity:  88 m/s

Average velocity:  50 m/s

initial velocity:  5.9 m/s

Final velocity:  94 m/s

Initial momentum:  3.6 kg m/s

Final momentum: 58 kg m/s

Explanation:

Acceleration = change in velocity / time

9.8 m/s² = Δv / 9.0 s

Δv = 88 m/s

Work = change in energy

Fd = ΔE

(6.0 N) d = 2700 J

d = 450 m

Average velocity = distance / time

v_avg = 450 m / 9.0 s

v_avg = 50 m/s

v − v₀ = 88 m/s

½ (v + v₀) = 50 m/s

Solving the system of equations:

v + v₀ = 100 m/s

2v = 188 m/s

v = 94 m/s

v₀ = 5.9 m/s

Use Newton's second law to find the mass:

F = ma

6.0 N = m (9.8 m/s²)

m = 0.61 kg

Find the momentums:

p₀ = (0.61 kg) (5.9 m/s) = 3.6 kg m/s

p = (0.61 kg) (94 m/s) = 58 kg m/s

6 0
3 years ago
As a freely falling object picks up downward speed, what happens to the power supplied by the gravitational force?
vitfil [10]

We will start from the definition of power in terms of the Force. Power could be described as the change of energy in an instant of time. Considering that Energy is the product between the Force and the distance traveled we would arrive at the expression

P = \frac{E}{t}

P = \frac{F*h}{t}

Here,

F = Force

h = Height

t = Time

As there is no external force, apart from the force of gravity, and this, is constant during the course of the object we will also have to be constant power and therefore this during its course will be the same. The correct answer is (1)

4 0
4 years ago
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