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o-na [289]
3 years ago
7

If two clay masses produce a gravitational force of 340N, what will be the force if the distance is divided by 3 and the mass of

one is divided by 2?
75.56 N
1530 N
510 N
56.67 N
Physics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

B: 1530 N

Explanation:

We know ghat formula for force of gravity is;

F_grav = G•m1•m2/d²

We are told that two clay masses produce a gravitational force of 340N.

Thus;

G•m1•m2/d² = 340

Now, if the distance is divided by 3 and the mass of one is divided by 2, we have;

F_g = (G × m1/2 × m2)/(d/3)²

Thus gives;

F_g = ½(G•m1•m2)/((1/9)d²)

Simplifying this gives;

F_g = (9/2)G•m1•m2/d²

From earlier, we saw that;

G•m1•m2/d² = 340

Thus;

F_g = (9/2) × 340

F_g = 1530 N

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Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the
Maksim231197 [3]

Answer:

<h3> 1.40625m/s²</h3>

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the  final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

Hence the acceleration due to gravity on the planet is  1.40625m/s²

6 0
3 years ago
What type of energy does the box have after it is done being pulled?
Ksivusya [100]
Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

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3 years ago
Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations
lapo4ka [179]

Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Explanation:

When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.

By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.

The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.

F= -kx, where,

F= restoring force

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k= constant related to the rigidity of the system.

Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.

5 0
3 years ago
Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

5 0
3 years ago
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