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horsena [70]
3 years ago
15

What is the relationship between mass and gravitational force between any two objects?

Physics
1 answer:
mamaluj [8]3 years ago
7 0
If two objects' mass increases, gravitational force pulls them down to the center more strongly because the greater the mass, the stronger the attraction is from the gravitational force from any two objects. An increase in then Earth's mass and increase in a man's mass would make gravitational force pull more strongly on the person because the Earth's gravitational force multiplied from the mass change and the man's greater mass pulls them down more quickly.
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A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board.
Vlada [557]

The velocity of the board relative to the ice is zero, since both are at rest.

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

Learn more about relative velocity at: brainly.com/question/24337516

#SPJ1

8 0
2 years ago
A rollerblader is blading along the sidewalk. Which forms of measurement would be the best to use to determine the rollerblader'
Greeley [361]
You would use distance an time formula to mathmaticly solve

3 0
2 years ago
A particle moves along a straight path through a displacement d = 2.5i + cj while a force F = 8.5i + -8.5j acts on it. The displ
Zanzabum

Answer:

Explanation:

Work is defined as the scalar product of force and distance

W=F•d

Given that

F = 8.5i + -8.5j. +×-=-

F=8.5i-8.5j

d = 2.5i + cj

If the work in the practice is zero, then W=0

therefore,

W=F•ds

0=F•ds

0=(8.5i -8.5j)•(2.5i + cj)

Note that

i.i=j.j=k.k=1

i.j=j.i=k.i=i.k=j.k=k.j=0

So applying this

0=(8.5i -8.5j)•(2.5i + cj)

0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)

0=21.25-8.5c

Therefore,

8.5c=21.25

c=21.25/8.5

c=2.5

7 0
3 years ago
which type of wave spreading do you think causes faster energy loss-two-dimensional or three-dimensional? explain.
sdas [7]
Three dimensional would loose faster 

3 0
3 years ago
Read 2 more answers
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
2 years ago
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