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3241004551 [841]
3 years ago
9

What model can you use to represent the forces acting on an object? How can it help you determine the net force on the object?

Physics
1 answer:
Nata [24]3 years ago
5 0

Answer:

Free body diagram

Explanation:

A free body diagram shows all the forces acting on an object. an example would be a box sitting on the floor.

Draw a square and you would have an arrow from the box pointed down to represent gravity pulling the box down and an arrow from the box pointing upwards to represent the normal force of the ground pushing back. in the scenario the two arrows would be equal length because the forces balance out since the box is motionless. in a situation where there is motion one arrow would be bigger such as if a box was falling. in this example it would have an arrow down to show gravity but no arrow up because it is in free fall. The sum of the forces which is represented by the arrows is what your net force is. The free body diagram helps you visualize all the forces acting on an object.

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Three individual point charges are placed at the following positions in the x-y plane:Q3= 5.0 nC at (x, y) = (0,0);Q2= -3.0 nC a
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Answer:

Explanation:

net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be  zero

if net  electric field due to charges Q₁ and Q₂  at origin is zero .

electric field due to Q₂

= 9 X 10⁹ X 3 x10⁹ / .04²

electric field due to Q₁

= 9 X 10⁹ X Q₁ / .02²

For equilibrium

9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²

Q₁  = 3 X10⁻⁹ x .02² / .04²

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The type of graph used to show how a part of something relates to the whole is which of the following?
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Read 2 more answers
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
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Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

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