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TEA [102]
3 years ago
10

Select all that apply. An acorn falls from a tree. Which of the following statements is true? The force of gravity is acting on

the acorn. The force of gravity is the only force acting on the acorn. A frictional drag force is acting on the acorn. The net force on the acorn is less than the force of gravity.

Physics
2 answers:
DedPeter [7]3 years ago
6 0

The net force on the acorn is less than the force of gravity.

Charra [1.4K]3 years ago
5 0

Answer:

The force of gravity is acting on the acorn.

A frictional drag force is acting on the acorn.

The net force on the acorn is less than the force of gravity.

Explanation:

The force of gravity is acting on the acorn is in vertically downwards direction.

A frictional drag force is acting on the acorn is in vertically upwards direction.

The net force on the acorn is given by

Net force = Force of gravity - Frictional drag force

It is always less than the force of gravity.

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Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

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Forces affect motions in living and nonliving things. In a human, swallowed food moves down the esophagus into the stomach, even
valina [46]

Answer:

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2 years ago
If an object is projected upward with an initial velocity of 127 ft per? sec, its height h after t seconds is h equals negative
Stolb23 [73]
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:

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What type of force are you exerting when you lie on a bed
zysi [14]

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Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
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