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SVEN [57.7K]
3 years ago
8

If forces acting on an object are unbalanced, which factor may result from an unbalanced force? The net force is negative. There

is zero net force. The forces cancel each other out. There is no change in motion.
Can you guys help me out, please??
Physics
2 answers:
NISA [10]3 years ago
8 0

Answer:

A.  The net force is negative

Explanation:

The net force is negative is the correct answer.  Just took the test on edge.

Oksi-84 [34.3K]3 years ago
6 0

Answer:the Forces cancel out each other

Explanation:the forces cancel out each other

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Explanation:

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3 years ago
Which statement is accurate about how the aurora borealis is formed?
g100num [7]

Answer:

When the electrons jump to a higher energy state, they release energy as electromagnetic radiation, light.

Explanation:

When the solar wind gets past the magnetic field and travels towards the Earth, it runs into the atmosphere. As the protons and electrons from the solar wind hit the particles in the Earth's atmosphere, they release energy – and this is what causes the northern lights.

7 0
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if the frequency of a sound wave in air remains constant, its energy can be varied by changing its what
Anastaziya [24]
Amplitude, is the answer to the question
6 0
3 years ago
Read 2 more answers
An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the
Alexxx [7]
(1) The position around equilibrium of an object in simple harmonic motion is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the motion
\omega is the angular frequency.

The velocity is the derivative of the position:
v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)
where 
v_0 = \omega A is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)
where
a_0=\omega^2 A is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
v_0 = \omega A = 4.3 m/s
a_0 = \omega^2 A = 0.65 m/s^2
From the first equation, we get
A= \frac{4.3 }{\omega} (1)
and if we substitute this into the second equation, we find the angular fequency
\omega=0.15 rad/s
while the amplitude is  (using (1)):
A=28.7 m


(b) We found in the previous step that the angular frequency of the motion is
\omega=0.15 rad/s
But the angular frequency is related to the period by
T= \frac{2 \pi}{\omega}
and so, the period is
T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s
5 0
3 years ago
A pencil is rolled off a table of height 1.25 m. If it has a horizontal speed of 3.1 m/s, how long does it take the pencil to re
Darya [45]
The horizontal speed has no effect on how long it takes to reach the ground.
A bullet shot from a gun and a bullet dropped from the front end of the gun
at the same time as the shot both hit the ground at the same time.

The number that counts is the height from which it fell . . . the 1.25 m.

I'll use this very useful formula:

       Distance of free fall,
       starting from rest            = (1/2) · (gravity) · (time)²

                                  1.25 m  =  (1/2) · (9.8 m/s²) · (time)²

Divide each side
by   4.9 m/s²  :          1.25 m / 4.9 m/s²  =  time²

                                          0.2551 sec²  =  time²

Square root each side:       0.505 sec  =  time

It looks like the correct choice is approximately 'A'. (rounded)
7 0
4 years ago
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