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tigry1 [53]
3 years ago
6

The force that contributed to the formation of planets, determines the motion of bodies in the solar system, and pulls objects t

o the center of Earth is called
Physics
1 answer:
kykrilka [37]3 years ago
6 0

Answer:

The force of gravity

Explanation:

Gravity was studied, by early scientists such as Copernicus and others, Galileo was the first to ensure that planets moved according to a physical equation that depended on a force that caused celestial bodies to move and interact with each other. But years later Newton based on studies conducted deciphering what Galileo assumed, he was able to find the equation of the force of gravity in any body in the universe. This equation depends on the masses of the two interacting bodies, the distance between them and a constant, which I call universal gravitation constant.

F_{g}=G*\frac{m_{1}*m_{2}}{r^2}

Fg = gravity force [N]

G =  universal gravitation constant = 6.67*10^(-11) [N*m^2/kg^2]

m1 = mass of the 1st body [kg]

m2 = mass of the 2nd body [kg]

r = distance between the bodies [meters]

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Properties of convex mirror
Ahat [919]

Answer:

The image result of an object reflected by a convex mirror is typically virtual, upright, and smaller. Discover how moving the object farther away from the mirror's surface affects the size of the virtual image formed behind the mirror

Explanation:

6 0
3 years ago
How is a controlled variable different from a responding variable?
Arisa [49]

Answer:

The answer is a controlled variable stays the same throughout an experiment, but a responding variable changes

Explanation:

just did it on apex

5 0
3 years ago
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0
3 years ago
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at
Ghella [55]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

mu=(M+m)v

85\times u=(85+130)\cdot 1.6

u=\frac{215}{85}\cdot 1.6

u=2.529\cdot 1.6=4.04 m/s  

4 0
3 years ago
A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo
Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be a m/s².

Given:

Mass of the block is, m=5\ kg

Force of pull is, F=40\ N

Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

F_{net}=F-f=40-8=32\ N

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0
3 years ago
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