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3 years ago

7

Do humans have a gravitational field if so, how strong

2 answers:

Kaylis [27]3 years ago

8 0

Yes, every living or nonliving object has its own gravitational field. However, a human's is so small it would be impossible to measure.

lilavasa [31]3 years ago

7 0

I'm pretty sure that you have mass. When it comes to gravity, anything that has mass behaves exactly like anything else that has mass. Any two of them are attracted to each other, with a force that's proportional to the product of their masses, and inversely proportional to the square of the distance between them.

Now, if you have mass, then you have weight ... it's the answer you give when somebody asks "How much do you weigh ?". That's the force of gravity between you and the Earth, pulling you toward the center of the Earth. But it doesn't stop there. There's also a force of gravity pulling the Earth toward the center of you. The strength of it is EQUAL to your weight.

Your weight on Earth is EQUAL to the Earth's weight on YOU !

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A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for

musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

$\Delta F= k \Delta x$

where

$\Delta F = F_2 - F_1$ is the change in the force on the spring, where

$F_1 = m_1 g = (0.300)(9.8)=2.94 N$ is the force applied when the hanging mass is

$m_1 = 0.300 kg$

$F_2 = m_2 g = (1.95)(9.8)=19.1 N$ is the force applied when the hanging mass is

$m_2 = 1.95 kg$

$\Delta x=x_2 -x_1$ is the change in extension of the spring, where

$x_1=0.200 m$ is the extension of the spring when the hanging mass is 0.300 kg

$x_2=0.750 m$ is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

$k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m$

2)

When the first mass is hanging on the spring, we have

$F_1 = k (x_1 - x_0)$

where:

$F_1$ is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

$x_1$ is the extension of the spring wrt its natural length

$x_0$ is the natural length of the spring (the unloaded length)

Here we have

$F_1=2.94 N$

k = 29.5 N/m

$x_1=0.200 m$

Solving for $x_0$ , we find:

$x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m$

5 0

3 years ago

Help me it’s for a test I have??

Volgvan

1. C
2. A
3. E
4. D
5. B
6. F
i might have 2 and 6 mixed up, not completely sure tho

4 0

3 years ago

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When an element tends to lose its valence electrons in chemical reactions, it behaves more like a _______ . When an element tend

vampirchik [111]

the answers are

metal

nonmetal

lower

higher

in the middle

7 0

4 years ago

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Most of the energy released by nuclear fission is in the form of: alpha rays, beta rays, or gamma rays?

Sedbober [7]

I think it is <span>Alpha rays.</span>

3 0

3 years ago

Which object is completely opaque?

Pavel [41]

Answer:

a

Explanation:

light can not pass through it completely

7 0

3 years ago