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Ainat [17]
3 years ago
12

Two objects have a force of gravity of 100 N. If the distance between both objects decreases by a factor of 7, while the masses

both decrease by a factor of 8. What is the new Fg?
Physics
1 answer:
Shkiper50 [21]3 years ago
8 0

The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is

<em>F</em> = <em>G M m</em> / <em>r</em> ²

Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is

<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²

<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)

<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²

In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.

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A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric
vichka [17]

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

4 0
3 years ago
Read 2 more answers
A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.
brilliants [131]
40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
7 0
4 years ago
In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.
JulijaS [17]

Answer:

more than

Explanation:

In a nuclear fusion reaction, the mass of the products is more than the mass of the reactants.

7 0
3 years ago
What is the force exerted on a charge of 2. 5 µC moving perpendicular through a magnetic field of 3. 0 × 102 T with a velocity o
stira [4]

The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

<h3>What is the force exerted on the charge?</h3>

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge

v = velocity of charge

q = charge

F=qvBsinα

F=2.5×10⁻⁶×5.0×10³×3.0×10²

F=37.5 N

Hence The force acting on the charge will be 3.75 N.

To learn more about the force acting on charge refer to ;

brainly.com/question/451411

F = q V B sinα

Where F is the force applied to a moving charge.

V = charge velocity

q stands for charge.

α = angle between V and B directions

As a result, the moving charge is subjected to a force of 3.75 Newton.

3 0
2 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
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