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2 years ago

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Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

re to double and the second mass m2 were to triple, what would happen to the magnitude of the force of attraction

1 answer:

Scilla [17]2 years ago

6 0

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>

Mass 1 = m₁
Mass 2 = m₂
Gravitational constant = G
Distance apart = r
Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>

Mass 1 = 2m₁
Mass 2 = 3m₂
Gravitational constant = G
Distance apart (r) = r
New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

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As a projectile falls, what happens to the components of velocity?

netineya [11]

Answer:

Option (c).

Explanation:

An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,

$v_x=v\cos\theta\ \text{and}\ v_y=v\sin\theta$

$\theta$ is the angle of projection

The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.

As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .

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3 years ago

An object is thrown upward from the top of a 144-foot building with an initial velocity of 128 feet per second. The height h of

ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

0 = -16t²+128t+144

16t²-128t-144= 0

$t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s$

Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0

3 years ago

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago

If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have?

Travka [436]

Answer:

Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.

Potential energy is when an object has some sort of potential eg. for motion such as in this example.

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3 years ago

The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the secon

Eva8 [605]

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

$P = h\rho g$

Where,

h = Height

$\rho$ = Density

g = Gravity

Our values are

$\rho = 1000kg/m^3 \rightarow$ density of water at normal conditions

h = 7.3m

$g = 9.8m/s^2$

PART A) Replacing these values to find the total pressure difference we have to

$P_1 = h_1 \rho g$

$P_1 = (7.3)(1000)(9.8)$

$P_1 = 71540Pa$

In this way the pressure change would be subject to

$\Delta = P_2-P_1$

$\Delta = 2.1*10^5Pa- 0.7154*10^5Pa$

$\Delta = 138460Pa$

$\Delta = 0.135Mpa$

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

$P_2 = H\rho g$

$2.1*10^5 = H (1000)(9.8)$

$H = 21.42m$

Therefore the high which could a faucet be before no water would flow from it is 21.42m

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3 years ago