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bearhunter [10]
3 years ago
15

Example of an unbalenced force

Physics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

Explanation:

unbalance force can be defined as the force that makes object slow up or down, in order words unbalanced force can be  defined as anything that can alter the state of motion of an object.

an example of an unbalanced force is likened to a situation of someone inside a lift going for a meeting at the top floor, the force which propels the lift upward is unbalanced, because it is greater than the force acting on the lift.

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An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?
Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0
4 years ago
Which of the following statements is NOT true?
Nataly [62]
I think it’s D but i’m not too sure
4 0
3 years ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
3 years ago
A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
wlad13 [49]
The car's average acceleration would be 1.25m/s^2 or 1.25meters/second/second. That looks to be the fourth one you've listed.
7 0
3 years ago
Read 2 more answers
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
Elenna [48]

The tension in the cable is 23.2 N

<h3>What is the tension in the string?</h3>

The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.

Tcosθ, is acting perpendicularly, Tcosθ = 0

Taking moments about the pivot:

Tsinθ * 2.2 = 4 * 9.8 * 0.7

Solving for θ;

θ = tan⁻¹(1.4/2.2) = 32.5°

T = 27.44/(sin 32.5 * 2.2)

T = 23.2 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

Learn more about moments of forces at: brainly.com/question/23826701

#SPJ1

3 0
2 years ago
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