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2 years ago

What are some forces that come in pairs?

Physics

1 answer:

Nikolay [14]2 years ago

3 0

Answer:

ACTION REACTION FORCES

Explanation:

When there is an action frce there will be a reaction force

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A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that

denpristay [2]

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

6 0

3 years ago

____ is the force that moves people to behave, think, and feel the way they do, resulting in behavior that is energized, directe

pogonyaev

Answer:

Motivation

Explanation:

Motivation is the force that directs one's behavior. This force is required for repeated actions. There are two types of motivation:

Intrinsic

This type of motivation is comes from the individual

Extrinsic

This type of motivation is comes from an external influence.

Both conscious and unconscious factors influence motivation.

5 0

2 years ago

Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orb

adell [148]

Answer:

F = G M m / R^2 gravitational force on planet of mass m.

None of these quantities change in the given hypothesis so

there will be no change in the orbit of mass m

3 0

2 years ago

Plzzz help me will mark brilliant

Setler79 [48]

Answer:

i hope it helped U

stay safe stay happy

3 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago