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2 years ago

Why does Newton's law of universal gravitation apply to all objects in the universe? Explain

2 answers:

5 0

Newton's law of Universal Gravitation tells that Force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

As there is mass for every object, and they are separated by certain distance (from 1 cm to more than hundreds of light years), they exerts Gravitational force on each other

Hope this helps!

SCORPION-xisa [38]2 years ago

3 0

Because all objects attract in proportion to the product of their masses.Gravitational interactions exist between all objects with an intensity that is directly proportional to the product of their masses.

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Two cars, with the same mass and traveling at the same speed, hit large trees head-on. One car has a rigid body that undergoes l

nordsb [41]

The crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

<h3>Newton's third law of motion</h3>

According to Newton's third law of motion, action and reaction are equal and opposite.

The car with rigid body will exert maximum force to the driver while the car with crumple zone will exert lesser force to the driver since the crumple zone will act as shock absorber, reducing the impact of the force on the driver.

Thus, the crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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8 0

1 year ago

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in

Ilya [14]

Answer:

F = 2π I R B

Explanation:

The magnetic force is described by the equation.

F = q v x B = i L x B

Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

F = i L B sin θ

In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

F = i L B

The loop length is

L = 2π R

F = i 2π R B

F = 2π I R B

The force is in the loop

8 0

2 years ago

Distance v. Time

Allushta [10]

Answer:

(m)

2 ( s )

Explanation:

ok...........

3 0

3 years ago

A WLAN transmitter that emites a 100 mW signal is connected to a cable with a 3dB loss. If the cable is connected to an antenna

irakobra [83]

Answer: 24 dBm

Explanation:

in the attachment

6 0

2 years ago

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

Learn more here:

brainly.com/question/20357188

8 0

2 years ago