Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
suppose a 1 square meter panel of colar cells has an efficiency of 20% and recieves the equivlent of 6 hours of direct sunlight per day. What average power, in watts, does the panel produce
Answer:
The average power per day is 1008 kW.
Explanation:
Solar constant = 1.4 kW/m2
efficiency = 20 %
area, a = 1 square meter
time = 6 hours
Energy falling on the panel in 6 hours = 1.4 x 6 x 3600 kJ
The output is
= 20 % of 1.4 x 6 x 3600
= 0.2 x 1.4 x 6 x 3600
= 6048 kJ
Average power per day is
= 6048/6 = 1008 kW
An object that gives off electromagnetic waves based on its temperature
demonstrates which phenomenon?
A. Emission spectra
B. Blackbody radiation
C. Quantum mechanics
D. Photoelectric effect
Answer:
B) Blackbody radiation
MARK THIS AS BRAINLIEST PLEAWESSS :)
Answer:
Black body radiation
Explanation:
A car is moving at 10 m/s on a horizontal road with friction on a dry day. The car can travel around a traffic circle with a minimum radius of 4.8 meters. It rains and the car around a traffic circle with a minimum radius of 11.8 meters. What is the percentage of the coefficient of static friction on the rainy day compared to the dry day
Answer:
[tex]\mu_w=86\%[/tex]
Explanation:
From the question we are told that:
Velocity on Dry road [tex]V_d=10m/s[/tex]
Radius Dry [tex]r_d=4.8[/tex]
Radius wet [tex]r_w=11.8[/tex]
Generally the equation for coefficient of static friction on the dry day is mathematically given by
[tex]\mu mg=\frac{mv^2}{r_d}[/tex]
[tex]\mu g=\frac{v^2}{r_d}[/tex]
[tex]\mu_d 9.8=\frac{10^2}{4.8}[/tex]
[tex]\mu_d=2.125[/tex]
Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by
[tex]\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}[/tex]
[tex]\frac{\mu_w}{2.125}=\frac{4.8}{11.8}[/tex]
[tex]\mu_w=0.86[/tex]
Therefore
[tex]\mu_w=86\%[/tex]
Vectors 퐴, 퐵and 퐶are added together. 퐴has a magnitude of 20.0 units and makes an angle of 60.0° counterclockwise from the negativex-axis. 퐵has a magnitude of 40.0 units and makes an angle of 30.0° counterclockwise from the positive x-axis.퐶has a magnitude of 35.0 units and makes an angle of 60.0° clockwise from the negative y-axis. Determine the magnitude of the resultant vector 퐴+퐵+퐶and its direction as an angle measured counterclockwise from the positive x-axis.
Answer:
Magnitude = 15.86 units
direction = 69 degree below negative X axis
Explanation:
A = 20 units at 60.0° counterclockwise from the negative x - axis
B = 40 units at 30.0° counterclockwise from the positive x - axis
C = 35 units at 60.0° clockwise from the negative y - axis
Write the vectors in the vector form
[tex]\overrightarrow{A} =20 (- cos 60 \widehat{i} - sin 60 \widehat{j})=- 10\widehat{i} - 17.3 \widehat{j}\\\\\overrightarrow{B} =40 (cos 30 \widehat{i} + sin 30 \widehat{j})= 34.6\widehat{i} +20 \widehat{j}\\\\\overrightarrow{C} =35 (- sin 60 \widehat{i} - cos 60 \widehat{j})=- 30.3\widehat{i} - 17.5 \widehat{j}\\\\Now\\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = (- 10 + 34.6 - 30.3) \widehat{i} + (-17.3 + 20-17.5)\widehat{j}\\\\[/tex]
[tex]\\\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = - 5.7\widehat{i} -14.8\widehat{j}[/tex]
The magnitude is given by
[tex]= \sqrt{5.7^2 + 14.8^2} = 15.86 units[/tex]
The direction is given by
[tex]tan\theta = \frac{- 14.8}{- 5.7}\\\\\theta= 69^o[/tex]
below negative X axis.
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? A. Potential energy B. Rotational energy C. Kinetic energy D. Translational energy
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.
Answer:
A. Potential energy
Explanation:
Potential energy can be thought of as stored energy, which has the potential of becoming kinetic energy once it has been released.
Work and the Dot Product
A variable 1D force acts on an object of mass 2 kg, which is initially moving at 5 m/s to the right (along the positive x direction). The net force is given by:
F x = 20x2-10 i Newtons x
The force acts on the object as it displaced from x = 1 m to x = 4 m .
a) Findthespeedoftheobjectatx=4m.
b) Is there a gain or loss in kinetic energy or no loss in kinetic energy in the
displacement of the object? Explain.
Answer:
a) v_f = 5,06 m/s, b) GAIN in kinetic energy.
Explanation:
a) For this exercise we will use the relationship between work and kinetic energy
W = ΔK
Work is defined by
W = F. d
bold indicates vectors
the displacement is
d = x_f - x₀
d = 4 -1
d = 3i m
we calculate
W = 20 10⁻² 3 i.i
let's remember that
i.i = j.j = 1
i.j = 0
W = 6.0 10⁻¹ J
we substitute in the first equation
W = K_f - K₀
W = ½ m (v_f ² -v₀²)
v_f ² = [tex]\frac{2W}{m} + v_o^2[/tex]
let's calculate
v_f ² = 2 6.0 10⁻¹ /2 + 5²
v_f = √25.6
v_f = 5.06 m / s
b) we can see that the speed at the end of the movement is greater than the initial speed, therefore there is a GAIN in kinetic energy.
does net force stay the same when a massless pulley is replaced by a pulley with mass
12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car comes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
What is the source of almost all energy on Earth?
Earth’s hot core
the Sun
stored carbon
moving water
Answer:
the sun
Explanation:
The sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
Source of energy on EarthThe Earth is one of the planets that make up the solar system. The Sun is the center of the universe and the Earth revolves round the sun.
The source of almost all energy on Earth is from the sun. The energy from the sun is callled solar energy.
This energy from the sun can be used by the planet to manufatcure its own food. The plants are consumed by animals to provide energy their metabolic activities.
Thus, we can conclude that the sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
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A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a. 50km/hr.
b. 10km/hr
Explanation:
Average speed, which is calculated by dividing the total distance travelled by the time interval as follows:
Average speed = total distance travelled ÷ time
Average velocity is calculated by dividing the total displacement by the time interval as follows:
Average velocity = change in displacement (∆x) ÷ time (t)
According to this question, a motorist travels due North at 90 km/h for 2 hours. She then changes direction and travels West at 60 km/for 1 hour.
Total distance of this journey is 90 + 60 = 150
Total time taken = 1 + 2 = 3hours
Average speed = 150/3
= 50km/hr.
b.) Average velocity = x2 - x1/t
Average velocity = 90 - 60/3
= 30/3
= 10km/hr
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.
Answer:
the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Explanation:
Given the data in the question;
Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s
Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s
the ratio Bonzo/mEnder of the masses of these two enemies = ?
Now, using the law of conservation of momentum.
momentum of both Bonzo and Ender are conserved
so
Initial momentum = final momentum
we have
0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]
[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex] ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]
we substitute
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -4.3 m/s / 1.1 m/s ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -3.9090 ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = 3.91
Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
the two factors that affect the amount of heat
Answer:
The two important factors that affect heat energy are specific heat and temperature. Specific heat is a heat-constant of a material per unit mass per degree of temperature change (in units of energy per mass and temperature), like Joules/Kg-°C .
Thank you.....
Have a good day.....
Two horizontal forces are acting on a box. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that = +5.9 N and the mass of the box is 3.6 kg. Find the magnitude and direction of when the acceleration of the box is +7.1 m/s^2.
Answer:
sorry I don't know I am only in 7th grade
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
D) 1.5032 x 10^-10 J
Explanation:
The rest energy of a proton, E₀, follows the equation:
E₀ = mp*rate²
Where mass of proton, mp = 1.6726 x 10^-27kg
rate = 2.9979 x 10^8 m/s (2.9979 x 10^9 m/s is not the speed light)
E₀ = 1.6726 x 10^-27kg * (2.9979 x 10^8 m/s)²
E₀ =1.5032 x 10^-10 J
Right answer is:
D) 1.5032 x 10^-10 JSuppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm
Answer:
0.035 J
Explanation:
Applying,
W = ke²/2.............. Equation 1
Where W = workdone by the stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = 2W/e²............... Equation 2
From the question
Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m
Substitute these values into equation 2
k = (2×2)/(0.15²)
k = 177.78 N/m
Hence, work need to stretch the spring from 33 cm to 35 cm
therefore,
e = 35-33 = 2 cm = 0.02 m
Substitute into equation 1
W = 177.78(0.02²)/2
W = 0.035 J
A spinning satellite begins to unfold two solar panels as shown. As the
panels extend from the satellite, what is the result?
Its angular momentum of a satellite will drop even as satellite panel stretch because of the reduction in angular velocity brought on by the drag effect.
Why then do stars spin more quickly during their demise?A portion of the star's mass is blown off during the supernovae that come before the creation of black holes, taking some of the star's total angular momentum with it.The leftover material sinks into the star's core while continuing to spin rapidly.
What is the name of the spinning effect?Magnus effect: When there is relative movement between both the spinning item and the fluid, a sideways force is generated on a rotating cylindrical or spherical object immersed in the fluid (liquid or gas).In honor of the German chemist and physicist H.G.
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Coherent light with wavelength 597 nm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.
For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Answer:
The required wavelength is 1.19 μm
Explanation:
In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.
[tex]y_m = \dfrac{m \lambda D}{d}[/tex]
where;
m = fringe order
d = slit separation
λ = wavelength
D = distance between screen to the source
For the first bright fringe, m = 1, and we make (d) the subject, we have:
[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]
[tex]d = \dfrac{ \lambda D}{y_1}[/tex]
replacing the value from the given question, we get:
[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]
In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:
[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
The value of m in the dark fringe first order = 0
∴
[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
making λ the subject of the formula, we have:
[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]
[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]
what happened on march 21 every year in the northern hemisphere
Answer:
B. The Spring equinox
Explanation:
The vernal equinox marks the moment the sun crosses the celestial equator. The vernal equinox happens on March 19, 20, or 21 every year in the Northern Hemisphere. In the Southern Hemisphere, this same event marks the beginning of fall. (Source: What Exactly Is The Spring Equinox? - Dictionary.com)
Hopefully this helps.
Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.
Answer:
Following are the responses to the given question:
Explanation:
The mass (mg) is downwards, the typical [tex]N_2[/tex] pressure is upward. This pushing force of [tex]\vec{F}_{3 \to 2}[/tex] is pushed forward by skater 3 on skater 2. (considered as rightward direction). The strength of the slater 2 in reply is the slater [tex]\vec{F}_{2\to 3}[/tex] Skater three moves to the left.
Its push force [tex]\vec{F}_{2\to1}[/tex] imparted to the skateboarder 2 in relation those above forces The force[tex]\vec{F}_ {1\to 2}[/tex] on skater 2 by skater 1 is as a response, to a left. Skater 2's free body system is as follows:
. How many meters away is a cliff if an echo is heard 0.5 s after the original sound? ( Assume that sound travels at 343 m/s on that day
Answer:
171.5 m
Explanation:
To find the distance, speed x time
342 x 0.5
171.5 m
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A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?
Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at 16 m/s while the other puck has a mass of 0.14 kg and a speed of 3.8 m/s. What is the velocity of the center of mass of the two pucks?
Answer:
10.66 m/s
Explanation:
Applying,
The law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').............. Equation 1
Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks
make V the subject of the equation
V = (mu+m'u')/(m+m').............. Equation 2
Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s
Substitute these values into equation 2
V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)
V = (2.88+0.532)/(0.32)
V = 3.412/0.32
V = 10.66 m/s
A soccer player kicks a ball. Why does the action force exerted by the player's foot cause a different motion than the reaction force?
The reaction force is greater than the action force.
The action force and the reaction force act on different objects.
The action force is greater than the reaction force.
The action force and the reaction force act in opposite directions.
Answer:
D because of POE
Explanation:
the reaction force is the ball exerting the same newtons of force back to your leg opposite of the ball, but the reaction force and the action force is never stronger than each-other. and action is only being done on the soccer ball, so process of elimination, the answer is D
Answer:
B
Explanation:
it can only have a different motion if it is acted upon a different ball
A motor has an output of 1000 watts. When the motor is working a full capacity, how much time will it require to lift a 50 Newton weight 100 meters?
The time required to lift the weight is 5 seconds.
What is time?Time is the measure of past or present events or occurrences. The S.I unit of time is seconds (s).
To calculate the time required to lift the weight, we use the formula below.
Formula:
P = Fd/t.................. Equation 1Where:
P = PowerF = Weightd = distance.t = timemake t the subject of the equation.
t = Fd/P................ Equation 2From the question,
Given:
F = 50 Nd = 100 mP = 1000 WSubstitute these values into equation 2
t = (50×100)/1000t = 5 seconds.Hence, The time required to lift the weight is 5 seconds.
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20 points, im begging for help‼️
How much capacitance is needed to
store 0.00100 J of energy when the
charge on the capacitor is
4.86 x 10-5 C?
[?] x 10?!F
Answer:
Capacitance= 1.18×10^-6
Answer: 1.18*10^-6
Explanation:
4. You currently have TWO 9 volt batteries connected to a 82 resistor. How
much current is flowing through the circuit? *
1.125 A
5 A
72 A
2.25 A
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Explanation:
Total voltage = 9 × 2 = 18v
Resistance = 82 Ω
Ohm's law::
V = IR
18v = 82 Ω × I
18v /82 /Ω = I
18/82 Ampere is the current
Give an example of a physical entity that is quantized. State specifically what the entity is and what the limits are on its values.
Answer:
A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...
Explanation:
One example of a physical entity that is quantized is:
The amount of money in your pocket.
The amount can't have any fraction of 1 cent.
Its value must be an integer-multiple of cents, or 0.01 dollar.
When it increases or decreases, it jumps from one integer number of cents to the next integer number. It doesn't "slide" from one to the next. It can never have a value between two integer numbers of cents.
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was 500 nm. What is the effective diameter of the circular laser aperture at the Maui ground station
This question is incomplete, the complete question is;
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.
What is the effective diameter of the circular laser aperture at the Maui ground station
Answer:
the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
Explanation:
Given the data in the question;
Separation between observer and point L = 354 km = 354000 m
Linear separation D = 9.1 m
wavelength λ = 500 nm = 500 × 10⁻⁹ m
Now, for small angles;
θ = D / L
θ = 9.1 m / 354000 m
θ = 2.57 × 10⁻⁵ rad
For a circular aperture;
sinθ = ( 1.22 × λ ) / d
for small angles;
θ = ( 1.22 × λ ) / d
so
θ = 2 × θ
θ = 2 × [( 1.22 × λ ) / d]
we substitute
2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]
2.57 × 10⁻⁵ = 0.00000122 / d
d = 0.00000122 / 2.57 × 10⁻⁵
d = 0.04747 m
d = ( 0.04747 × 100 )m
d = 4.747 cm
Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
Suppose the Earth were squeezed down to the size of a small mountain. The mass of the Earth would not change, just its volume and radius. If you were now standing on the surface, would the gravitational force on you be greater than, less than or the same as before the Earth was squeezed?
Explanation:
Much greater. Gravitational force depends on the mass and separation distance. Shrinking the earth means the mass remains the same while the radius gets smaller. Since gravitational force is inversely proportional to the square of the separation distance, as shown by Newton's universal gravitational law
[tex]\:\:\:F_G = G \dfrac{mM_E}{R^2}[/tex]
shrinking the radius even by a factor of 10 will cause your weight, which also happens to be the gravitational force of the earth on you, to be 100 times more.
If the mass of the Earth remains the same and only its volume and radius are decreased, then the average density of the Earth would increase significantly.
What is gravitational force?Gravity, also known as gravitational force, pulls objects with mass towards each other. We frequently consider the force of gravity from Earth.
If the Earth's mass remains constant while its volume and radius are reduced, the average density of the Earth increases significantly.
This is due to the fact that the mass has been compressed into a much smaller volume.
As a result, standing on the surface, the gravitational force on you would be greater than before the Earth was compressed to the size of a small mountain.
When the Earth is compressed, the distance between you and the center of the Earth decreases while the mass remains constant.
Thus, your gravitational force increases.
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