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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pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts
Answer:
C. 6.9 watts
Explanation:
Power = work/time
if work = force×distance...
Then... power= (force×distance)/time
Power = (20×12)/35
= 6.9 watts
A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod.
a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. (Hint: Use the power series expansion for ln(1+x).)
b) Use Fx=−dU/dx to find the magnitude of the gravitational force exerted on the sphere by the rod.
Answer:
Explanation:
From the given information:
a)
Let's have an imaginary view of the rod located at a given distance r from he the mass (m) of the sphere.
Then the equation for the potential energy as related to the small area of the dr of the rod can be computed as:
[tex]dU = -\dfrac{GMm}{L}*\dfrac{dr}{r}[/tex]
where,
G = gravitational constant
[tex]U = - \int^{x+L}_{x}Gm\dfrac{M}{L}*\dfrac{dr}{r}[/tex]
[tex]U = - \dfrac{GMm}{L}\int^{x+L}_{x}\dfrac{dr}{r}[/tex]
By taking the integral within the limit
[tex]U = - \dfrac{GMm}{L} \Big[In \ r\Big]^{x+L}_{x}[/tex]
[tex]\mathbf{\implies - \dfrac{GmM}{L} In \Big(\dfrac{{x+L}}{{x}}\Big)}[/tex]
b)
By using [tex]F= -\dfrac{dU}{dx}[/tex], the magnitude of the gravitational force can be determined as follows:
Here, we have:
[tex]F = -\dfrac{d}{dx} \Big [\dfrac{-GmM}{L}In(\dfrac{x+l}{x}) \Big ] \\ \\ = \dfrac{GmM}{L}\times \dfrac{x}{x+L}\times (0-\dfrac{L}{x^2}) \\ \\ By \ solving \\ \\ \mathbf{ =-\dfrac{GmM}{x(x+L)}}[/tex]
From above, the negative sign indicates an attractive force
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.
A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 30 A current. What acceleration (in g's) does this interaction give the airplane?
Answer:
[tex]a=0.2*10^{-5}g[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=180=>0.18kg[/tex]
Charge [tex]Q=18mC=18*10^-^3C[/tex]
Velocity [tex]v=2.2m/s[/tex]
Length of Wire [tex]L=8.6cm=>0.086[/tex]
Current [tex]I=30A[/tex]
Generally the equation for Magnetic Field of Wire B is mathematically given by
[tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]
[tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]
[tex]B=6.978*10^{-5}T[/tex]
Generally the equation for Force on the plane F is mathematically given by
[tex]F=qvB[/tex]
Therefore
[tex]ma=qvB[/tex]
[tex]a=\frac{qvB}{m}[/tex]
[tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]
[tex]a=2.37*10^{-5}[/tex]
Therefore in Terms of g's
[tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]
[tex]a=0.2*10^{-5}g[/tex]
Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?
a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.
Answer:
Option (a).
Explanation:
Let the angular velocity is w.
The centripetal acceleration is given by
[tex]a = r w^2[/tex]
where, r is the distance between the axle and the spoke.
So, more is the distance more is the centripetal acceleration.
(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.
The statement is true.
(b) The direction of centripetal acceleration is always towards the center, so the statement is false.
(c) It is false.
(d) It is false.
Option (a) is correct.
Airplane lift is achieved when air pressure on the bottom of its wings is
A) greater than pressure on top.
B) less than pressure on top.
C) the same as pressure on top.
Answer:
C the same as pressure on top
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]
define emperical formula and what is the dimensional formula of force and energy
Answer:
An empirical formula represents the simplest whole number ratio of various atoms present in a compound.The dimensional formula of force is [[tex]MLT^{-2}[/tex]]The dimensional formula of energy is [[tex]ML^{2} T^{-2}[/tex]]3. Four charges having charge q are placed at the corners of a square with sides of length L. What is the magnitude of the force acting on any of the charges
Answer:
Fr = 1.91 * 9*10⁹*q²/L²
Explanation:
Let´s say that the corners of the square are A B C and D
We are going to find out the force on the charge placed on B ( the charge placed in the upper right corner.
As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.
Force due to charge placed in A
module Fₓ = K* q² / L² in the direction of x
Force due to charge placed in C
module Fy = K* q²/L² in the direction of y
Force due to the charge placed in D
That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.
d² = L² + L² = 2*L²
d = √2 * L
The module of the force due to charge place in D
F₄₅ = K*q²/ 2*L²
To get the force we need to add first Fₓ and Fy
Fx + Fy = F₁
module of F₁ = √ Fx² + Fy² the direction will be the same as the diagonal of the square then:
F₁ = √ ( K* q²/L² )² + ( K* q²/L² )²
F₁ = √ 2 * K*q²/L²
And now we add forces F₁ and F₄₅ to get the net force Fr on charge in point B.
The direction of Fr is the direction of the diagonal and is of rejection
the module is
Fr = F₁ * F₄₅
Fr = √ 2 * K*q²/L² + K*q²/ 2*L²
Fr = ( √ 2 + 0,5 ) * K*q² /L²
K = 9*10⁹ Nm²C²
Fr = 1.91 * 9*10⁹*q²/L²
We don´t know units of L and q
Explain why energy cannot escape from the room by conduction
Answer: Heat energy transfer by conduction, convection and radiation
Heat energy is a very difficult energy to store as it can transfer in three different ways from warm surroundings to cooler surroundings. The three processes are conduction, convection or radiation.
Understanding energy, how it is transferred and how the amount of energy that is usefully transferred can be used to measure the efficiency is very important to physics and to the world.
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.
Learn more about time here: https://brainly.com/question/4931057
what does Newton's third law ? Describe.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.
Hope it helps
Answer:
Newton's third law state that every action there is equal but opposite reactions
hope this will help you more
If a small child swallowed a safety pin, why
would an X-ray photograph clearly show the
location of the pin?
Answer:
yes
Explanation:
it is in the body system
Answer:
it would show clearly because it is a metal piece in the body.
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees rest on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero.
Required:
Do the bees have the same or opposite signs of charge?
Answer:
the charge of the bees must be of the same sign
Explanation:
The electric field is given by the relation
E = k q / r²
This electric field has outgoing direction if the charge is positive and incoming towards the charge if it is negative.
The force generated by this field on a test charge is
F = q E
Since the charge is a scalar, the direction of the force is the same as the electric field.
In this case the two flowers are at a certain distance and the two charged bees land on them, so the force on a test charge is the vector sum of the force that each bee creates, so that this force is subtracted from the two bees must have the same charge sign.
The force created by the bee on the left goes to the right and the force created by the bee on the right goes to the left, so the forces are subtracted,
Consequently the charge of the bees must be of the same sign
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Two forces and are applied to an object whose mass is 11.8 kg. The larger force is . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when points due east and points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of and (b) the magnitude of .
Can you please fill in whatever goes in the blanks ?
Without them, the question makes no sense and has no answer.
Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .
identify the word being referred to choose your answer from the words below
Answer:
1:Rotation
2:Axis
3:Aphelion
4:orbit
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?
Answer:
The answer is "Option D".
Explanation:
Please find the complete question in the attached file.
As plate separation increased to 2d the capacitance get halred but the change remain same
[tex]\therefore V=\frac{Q}{C}[/tex]
The voltage doubles are now electric field remain same because both the distance and voltage get doubled.
[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]
So,
[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?
a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min
Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]
let's calculate
v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]
v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
How can magnetic levitation be improved?
2. Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave detector. In an experiment, the beat frequency is 170MHz . One microwave generator is set to emit microwaves with a wavelength of 1.410cm.
If the second generator emits the longer wavelength, what is that wavelength?
Answer:
The wavelength of the wave is 1.419 cm.
Explanation:
wavelength emitted by the one micro wave = 1.41 cm
beat frequency = 170 M Hz
Let the frequency of the wave is f.
[tex]f=\frac{c}{\lambda }\\\\f =\frac{3\times 10^8}{0.0141}\\\\f = 2.13\times 10^{10} Hz[/tex]
Let the frequency of the other wave is f'.
[tex]f' = 2.13\times 10^{10}-170\times10^6\\\\f' = 21130\times 10^6 Hz[/tex]
The wavelength is given by
[tex]f' = 21130\times 10^6 Hz\\\lambda = \frac{3\times 10^8}{21130\times 10^6}\\\\\lambda = 0.01419 m = 1.419 cm[/tex]
You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
A kangaroo kicks downward with a 1000N force. According to Newton's Law the kangaroo is propelled into the air by:
A) gravitational force
B) his muscles
C) The earth
D) wallabies
Explanation:
Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.
This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
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:
djfjci3jsjdjdjdjdjddndn
ds
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
I NEED THE ANSWER QUICK PLEASEE