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LekaFEV [45]
3 years ago
14

If all of the forces acting on an object are balanced, then:

Physics
1 answer:
kondaur [170]3 years ago
3 0

A. the direction the object is moving in will not change.

B. the acceleration of the object will be 0 m/s2

Explanation:

We can answer this problem by using Newton's second law of motion:

\sum F = ma

where

\sum F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, all the forces acting on the object are balanced, therefore the net force is zero:

\sum F=0

which means that also the acceleration is zero:

a=0

Acceleration is equal to the rate of change of velocity: therefore, zero acceleration means that the velocity of the object does not change. We can now analyze the given statements:

A. the direction the object is moving in will not change.  --> TRUE, because the velocity is not changing.

B. the acceleration of the object will be 0 m/s2  --> TRUE, as we stated above

c. the object will not be in motion.  --> FALSE: we just know that its velocity is constant, but it can be different from zero

D. the velocity of the object will be 0 m/s. --> FALSE, for the same reason stated in C

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

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A light bulb whose resistance is 240 ohms is connected to a 120 voltage source. What is the current through the bulb?
Sedaia [141]

Answer:

0.5A

Explanation:

Using R = \frac{V}{I},

R is the resistance (in Ohms)

V is the voltage (in V)

I is the current (in A)

240=\frac{120}{I}

I = 0.5A

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Which of these values is equivalent to the change in momentum of
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<span>A. An impulse of a force changes the momentum of a body and has the same units and dimensions as momentum.</span>
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An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon
erastova [34]

The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ_{1} = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ_{2} = Adjacent side 2 / Hypotenuse

cos θ_{2} = 4.98 / 8.5 = 0.59

θ_{2} = cos^{-1} ( 0.59 )

θ_{2} = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

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3 0
10 months ago
a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

5 0
3 years ago
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