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kramer
3 years ago
7

A rock is stuck deep in the dirt. When you pull on the rock to remove it, what type of force are you exerting?

Physics
2 answers:
aleksandrvk [35]3 years ago
4 0
It is Tension as the other 3 answer choices would not make sense. Compression would mean you are pressing the rock on both sides or in this case, pushing it into the dirt. It can't be nuclear force as you are pulling out a rock. Air resistance would not make sense either as there is no air involved in the scenario at all.
Mazyrski [523]3 years ago
4 0

Answer:

C). Tension Force

Explanation:

As we know that the different forces will act at different situations

A) Compression : here when compression occurs then the force is applied along the length of the object due to which its length will contract and it will apply force linearly.

B) Strong Nuclear Force : This only occurs are nuclear level when nuclei comes closer to each other then only this force will exert and it will be the strongest force.

C). Tension : This force will occur when we pull an object and there is no change in the length of the object. So tension force will be the force when there is a pull.

D) Air resistance : this force is due to opposite force of air molecules due to which it resist the motion of object

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The change in momentum of an object is equal to the ____________ that acts on it.
meriva

Answer : The change in momentum of an object is equal to the impulse that acts on it.

Explanation :

Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.

The formula of change in momentum is,

\Delta p=m\times \Delta v

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of  an object.

J=F\times \Delta t

Proof :

J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p

Hence, the change in momentum of an object is equal to the impulse that acts on it.

3 0
3 years ago
Read 2 more answers
A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

4 0
3 years ago
Read 2 more answers
If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
2 years ago
A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T .
kupik [55]

Answer:

t = 3/2T

To find how long it takes to cover a total distance of 6A, we need to find the time it takes to cover a distance A then multiply by 6.

The step to the solution is given below in the attachment.

Explanation:

Thank you for reading

4 0
2 years ago
What was an ill effect of the industrial revolution?
Digiron [165]

Answer:

answer is option (c) child labour

6 0
3 years ago
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