State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitational force.

Answers

Answer 1

Answer: F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

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A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. The coefficient of static friction is mu_s, and the acceleration due to gravity is g. Find the total work W done on the car by the external forces.
In a series circuit, each circuit element has the same:a.currentc.capacitanceb.voltaged.resistance
If gravity did not affect the pain of a horizontally thrown ball the ball would
3. An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph.Answer the following questions:a. What is the period and frequency of the pendulum’s motion?b. How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?c. What is the acceleration due to gravity on the surface of the planet in m/s2? Determine the number of g-forces. Show any necessary calculations.Answer:
Anyone please??? will rate brainless and would help a lot please

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance between the photometer to lamp 1 is 400mm, the distance between the photometer to lamp 2 is 200 mm, and the intensity of lamp 2 is known to be 15 candelas, what is the intensity to lamp 1?

Answers

Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 candela

In scientific terms, how is power related to work? A.Power is the ability to do work.

B.Power is the force of work over a certain distance.

C.Power is the rate at which work is accomplished.

D.Power is the energy supplied in a force.

Answers

C. the rate at which work is done. hope i helped!

Given: Q1 = 3.0 × 10-5 CQ2 = 4.0 × 10-5 C
K = 9.00 × 109 N ×
r = 3.0 m, determine F.

_____ N

Answers

The force is obtained from the charges Q₁ = 3×10⁻⁵ C and Q₂ = 4×10⁻⁵ C and distance r = 3 m is 1.2 N.

What is Coloumb's law?

Coloumb's law is defined as the force of attraction and repulsion between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

The attractive and repulsive force, F∝ (q₁×q₂) / r². It is also known as F = k (q₁×q₂) / r², where k is the proportional constant. The unit of force is newton N.

From the given,

Q₁ = 3×10⁻⁵ C

Q₂ = 4×10⁻⁵ C

r = 3m

k = 9×10⁹ Nm²C⁻²

The attractive and repulsive force,

F = k (q₁×q₂) / r²

= (9×10⁹× 3×10⁻⁵×4×10⁻⁵) / (3)²

= 12×10⁻¹

= 1.2 N

Force = 1.2 N

Thus, the force of repulsion and attraction of given charges is 1.2 N.

To know more about Coloumb's law:

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F=KQ1Q2/R2
F=9 x 10^9 x 3 x 10^-5 x 4 x 10^-5 / 9
F=12 x 10^-1
F=1.2 N

A scientist is conducting research to determine how much pressure a dam can withstand. Why are good observations important to his research?

Answers

The things that a scientist should consider while observing the force is the environmental conditions,the force that is expected to act on the dam, the means to contain that force, and compare different types of designs in accordance with the location of the dam

What is the current when the resistance is 5 ohms and the voltage is 10 volts?A. 10 – 5 = 5 amps

B. 10 ÷ 5 = 2 amps

C. 10 × 5 = 50 amps

D. 10 + 5 = 15 amps

Answers

Ohm's Law states V = IR
So,
I = V/R
The answer is B. 10/5=2 amps

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. then the girl pulls on the rope, exerting a 5.76 N force on the sled, pulling it toward her. how far from the girl's original position do they meet?

Answers

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=(F)/(m)$ . Now $a_(girl)=(5.76)/(42.3)=0.14 (m/s^(2))$ and $a_(sled)=(5.76)/(7.93)=0.73(m/s^(2))$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=(1)/(2)*a_(girl)*t^(2)$ and $15.0-x=(1)/(2)*a_(sled)*t^(2)$ , solving for the time we get: $t=\sqrt{(2x)/(a_(girl) ) }$ and $t=\sqrt{(2*(15.0-x))/(a_(sled) ) }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{(2x)/(a_(girl))} = \sqrt{(2*(15.0-x))/(a_(sled) )$ . Finally we get: $(x)/(a_(girl) )=((15.0-x))/(a_(sled) )$ and replacing the values we have got: $(x)/(0.14) =((15.0-x))/(0.73)$ so $5.33*x=15-x$ so x=2.37 (meters).

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