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Nana76 [90]
3 years ago
10

Which pair of objects will have the least gravitational attraction to each other

Physics
1 answer:
polet [3.4K]3 years ago
5 0

Hey i am here to help you.

A: 2 concrete blocks 6cm apart will have the least gravitational attraction to each other.

I believe that this answer was helpful.

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Assume you are given an int variable named nElements and a two-dimensional array that has been created and assigned to a2d. Writ
Alex777 [14]

Answer:

Explained

Explanation:

public int dimension(int [][]a2d,int nElements)

{

int count = 0;

for(int i = 0;i < a2d.length ; i++)

{

count = count + a2d[i].length;

}

return count;

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3 0
3 years ago
7. How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a
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The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
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3 years ago
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Explain why it is important to keep all variables that might affect the dependent variable, other than the manipulated variable
Inga [223]
<span>You have multiple confounding variables, you cannot accurately conclude the relationship between the manipulated and dependent variable because the other variables that are not controlled for could be the reason for seeing a certain change</span>
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3 years ago
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A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in
timama [110]

Answer:

KE2/KE1=0.71

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

V2 = \frac{m1}{m1+m2}*V1

The kinetic energies are:

KE1=1/2*m1*V1^2

KE2 = 1/2*(m1+m2)*V2^2

KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2

Simplifying:

KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2

The ratio will be:

KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}

KE2/KE1=0.71

5 0
3 years ago
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