Answer:
1 m/s²
Explanation:
a = v² / r
a = ω² r
a = (2π rad / 2π s)² (1 m)
a = 1 m/s²
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring constant k.
Answer:
k = 0.5 MN/m
Explanation:
Mass of the railcar, m = 5000 kg
Speed of the rail car, v = 1 m/s
The Kinetic energy(KE) of the railcar is given by the equation:
KE = 0.5 mv²
KE = 0.5 * 5000 * 1²
KE = 2500 J
The spring's compression, x = 0.1 m
The potential energy(PE) stored in the spring is given by the equation:
PE = 0.5kx²
PE = 0.5 * k * 0.1²
PE = 0.005k
According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring
KE = PE
2500 = 0.005k
k = 2500/0.005
k = 500000 N/m
k = 0.5 MN/m
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoir
Answer:
Q₂ = 5833.33 J
Explanation:
First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:
η = W/Q₁
where,
η = efficiency of engine = 30% = 0.3
W = Work done by engine = 2500 J
Q₁ = Heat supplied to the engine = ?
Therefore,
0.3 = 2500 J/Q₁
Q₁ = 2500 J/0.3
Q₁ = 8333.33 J
Now, we find the heat discharged to lower temperature reservoir by using the formula of work:
W = Q₁ - Q₂
Q₂ = Q₁ - W
where,
Q₂ = Heat discharged to the lower temperature reservoir = ?
Therefore,
Q₂ = 8333.33 J - 2500 J
Q₂ = 5833.33 J
A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz
Answer:
The length = 27.52m
Explanation:
v=f x wavelength
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the initial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?
Answer:
Pls see attached file
Explanation:
Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
A 240 m long segment of wire is hanging between 2 transmission lines. What is the total magnetic force only (ignore gravitational force) on this segment of wire if the current in the wire is 500 A and the field strength is 3e-5 T
Answer:
7.2 N
Explanation:
length of wire L = 240 m
current I = 500 A
field strength B = 3 x 10^-5 T
magnetic force on a current carrying conductor F is given as
F = BILsin∅
The wires are perpendicular with field therefore sin∅ = sin 90° = 1
therefore,
F = BIL = 3 x 10^-5 X 500 X 240 = 3.6 N
If the wire exists between this two transmission lines, then total magnetic force on the wire = 2 x 3.6 = 7.2 N
(9) A dancer spins at 72rpm about an axis through the center with arms outstretched. The dancer mass=90kg, model the head as a uniform sphere r=8.0cm Model the trunk and legs as uniform solid cylinder r=12.0cm, arms as slender rods L=60.0cm. On the average the head=7%, both arms together=13% and trunk and legs=80% of body mass. Find the K.E. of the dancer note: arms are I around an axis for slender rod: 1/3ML^2 each arm note2: Moments of Inertia are summable, i.e. I total = I1 + I2 + I 3 +...
Answer:
Explanation:
ω = angular velocity = 2π n = 2π x 72 / 60
= 7.536 rad /s
mass of head = 90 x .07 = 6.3 kg
moment of inertia of head = 2 /5 m R²
= .4 x 6.3 x .08²
= .016128 kg m²
moment of inertia of trunk + legs
= 1/2 m R²
= .5 x .8 x 90 x .12²
= .5184 kg m²
moment of inertia of arms
= 1/3 m L²
= 1 / 3 x .13 x 90 x .60²
= 1.404 kg m²
Total moment of inertia
I = 1.938 kg m²
kinetic energy = 1/ 2 I ω ²
where I is moment of inertia and ω is angular velocity
= .5 x 1.9338 x 7.536²
=55 J approx .
The definition of kinetic energy and moment of inertia allows finding the result for the kinetic energy of the dancer with arms extended is:
The kinetic energy is: KE = 56.3J
Given parameters
Frequency is: f = 72 rad / min (1 min / 60s) = 1.2 rad / s The mass is m = 90 k Head radius is r₁ = 8.0 cm = 0.08 m The leg and trunk are cylinders of radius r₂ = 12.0 cm = 0.12 m The length of the arms, approximated as rods L = 60.0 cm = 0.600 m The dancer spins with outstretched arms The percentage of the mass is:* Head 7%
* Arms 13%
* Trunk and legs 80%
To find.
Kinetic energy
The kinetic energy is the energy due to the movement, in this case the movement is rotational, therefore the expression is:
KE = ½ I w²
The angular velocity is related to the frequency.
w = 2π f
w = 2π 1.2
w = 7.540 s
The moment of inertia is a scalar, therefore it is a quantity that can be added, the total moment of inertia of the dancer is the sum of the moments of inertia of each part with respect to the axis of rotation of the person.
[tex]I_{toal}= I_{head}+ I_{trunk}+I_{arms}[/tex]
The moments of inertia with respect to the centers of mass are tabulated.
Sphere I = 2/5 mr²
Cylinder I = ½ m r²
Rod I = ⅓ m r²
The axis of rotation of the head and the trunk are in the axis of rotation of the person, therefore their moment of inertia is those corresponding to the center of mass.
At the end of the arms it is at a distance of D = [tex]\frac{r_2}{2}[/tex] from the axis of rotation of the dancer, therefore to find the moment of inertia we must use the theorem of parallel axes, see attached.
[tex]I_{arms} = I_{cm} + M D^2[/tex]
[tex]I_{arms}[/tex] = ⅓ m L² + m D²
[tex]I_{arms} = m_{arms} ( \frac{L^2}{3} + D^2)[/tex]
The masses of each part of the body are:
[tex]m_{head}[/tex] = m 0.07
[tex]m_{trunk}[/tex] = m 0.80
[tex]m_{arms}[/tex] = m 0.13
Let's find the total moment of inertia.
[tex]I_{total} = \frac{2}{5} \ 0.07m \ r_1^2 + 0.13m\ ( \frac{L^2}{3} + D^2) + \frac{1}{2} \ 0.80m \ r_2^2[/tex]
[tex]I_{total} = m ( 0.028 \ r_1^2 + 0.13 (\frac{L^2}{3} + D^2) + 0.40 \ r_2^2}[/tex]
Let's calculate.
[tex]I_{total} = 90\ ( 0.028 \ 0.08^2 + 0.13\ ( \frac{0.6^2}{3} + 0.06^2) + 0.40 \ 0.12^2 )[/tex]
[tex]I_{total}= 90 \ 0.0220 \\ \\I_{total} = 1.9806 \ Kg m^2[/tex]
we substitute in the kinetic energy
KE = ½ 1.9806 7,540²
KE = 56.3 J
In conclusion using the definition of kinetic energy and moment of inertia we can find the result for the kinetic energy of the dancer with the extended arms is:
The kinetic energy is: KE = 56.3 J
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A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.
Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?
Answer:
a
[tex]B = 0.0533 \ T[/tex]
b
[tex]B = 0.04 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]
The outer radius is [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]
The nu umber of turns is [tex]N = 960[/tex]
The current it is carrying is [tex]I = 2. 5 A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value
[tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]
And the given distance where the magnetic field is felt is r = 0.9 cm = 0.009 m
Now substituting values
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]
[tex]B = 0.0533 \ T[/tex]
Fro the second question the distance of the position considered from the center is r = 1.2 cm = 0.012 m
So the magnetic field is
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]
[tex]B = 0.04 \ T[/tex]
The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.
The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.
The given parameters;
radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 AThe magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]
The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]
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A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
Two children are balanced on a seesaw, but one child weighs twice as much as the other child. The heavier child is sitting half as far from the pivot as is the lighter child. Since the seesaw is balanced, the heavier child is exerting on the seesaw:_______.
a. a force that is less than the force the lighter child is exerting.
b. a force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Answer:
B. A force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Explanation:
If they are sitting at the same distance away from the pivot yet the seesaw is balanced, the only conclusion is the heavier child is exerting a lower force. This causes the pivot exertion and balances to be equal. The equilibrium of the pivot-seesaw is not affected by the weight because of force exertion.
How much electricity is in human body?
Answer:
100 on average
Explanation:
The human body is a physical substance from where a human can breathe, drink, eat, walk, etc, and consists of various cells. A human being cannot live with their functions and when functions stop working that means a human being is dead.
Therefore, in the human body, there is also electricity that contains 100 watts on average.
So, According to the given situation, the correct answer is 100 on average.
Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?
Answer:
- maximum displacement = 3.00nm
- λ = 1.79m
- f = 286.47 s^-1
Explanation:
You have the following equation for a sound wave:
[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex] (1)
The general form of the equation of a sound wave can be expressed as the following formula:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
A: amplitude of the wave = 3.00nm
k: wave number = 3.50m^-1
w: angular frequency = 1,800s^-1
- The maximum displacement of the wave is given by the amplitude of the wave, then you have:
maximum displacement = A = 3.00nm
- The wavelength is given by :
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]
The values for the wavelength is 1.79m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]
The frequency is 286.47s-1
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
(a) max. height = 3.641 m
(b) flight time = 1.723 s
(c) horizontal range = 31.235 m
(d) impact velocity = 20 m/s
Above values have been given to third decimal. Adjust significant figures to suit accuracy required.
Explanation:
This problem requires the use of kinematics equations
v1^2-v0^2=2aS .............(1)
v1.t + at^2/2 = S ............(2)
where
v0=initial velocity
v1=final velocity
a=acceleration
S=distance travelled
SI units and degrees will be used throughout
Let
theta = angle of elevation = 25 degrees above horizontal
v=initial velocity at 25 degrees elevation in m/s
a = g = -9.81 = acceleration due to gravity (downwards)
(a) Maximum height
Consider vertical direction,
v0 = v sin(theta) = 8.452 m/s
To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,
solve for S in equation (1)
v1^2 - v0^2 = 2aS
S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s
(b) total flight time
We solve for the time t when the vertical height of the object is AGAIN = 0.
Using equation (2) for vertical direction,
v0*t + at^2/2 = S substitute values
8.452*t + (-9.81)t^2 = 3.641
Solve for t in the above quadratic equation to get t=0, or t=1.723 s.
So time for the flight = 1.723 s
(c) Horiontal range
We know the horizontal velocity is constant (neglect air resistance) at
vh = v*cos(theta) = 25*cos(25) = 18.126 m/s
Time of flight = 1.723 s
Horizontal range = 18.126 m/s * 1.723 s = 31.235 m
(d) Magnitude of object on hitting ground, Vfinal
By symmetry of the trajectory, Vfinal = v = 20, or
Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s
What force is required so that a particle of mass m has the position function r(t) = t3 i + 7t2 j + t3 k?
Answer:
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k\\F(t)=\,(6\,m\,t,14\,m,6\,m\,t)[/tex]
Explanation:
Recall that force is defined as mass times acceleration, and acceleration is the second derivative with respect to time of the position. Since the position comes in terms of time, and with separate functions for each component in the three dimensional space, we first calculate the velocity (with the first derivative, and then the acceleration as the second derivative:
[tex]r(t)=t^3\,\hat i+7\,t^2\,\hat j+t^3\,\hat k\\v(t)=3\,t^2\,\hat i+14\,t\,\hat j+3\,t^2\,\hat k\\a(t)=6\,t\,\hat i+14\,\hat j+6\,t\,\hat k[/tex]
Therefore, the force will be given by the product of this acceleration times the mass "m":
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k[/tex]
If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 will not be "picked up" by this pickup?
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Answer:
b
Explanation:
because:/
The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of Earth towards the sun?
Answer:
ar = 5.86*10^-3 m/s^2
Explanation:
In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.
You use the following formula:
[tex]v=\sqrt{\frac{GM_s}{r}}[/tex] (1)
G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2
Ms: Sun's mass = 1.98*10^30 kg
r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m
Furthermore, you take into account that the radial acceleration is given by:
[tex]a_r=\frac{v^2}{r}[/tex] (2)
You replace the equation (1) into the equation (2) and replace the values of all parameters:
[tex]a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}[/tex]
The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2
A bicycle tire pump has a piston with area 0.43 In2. If a person exerts a force of 16 lb on the piston while Inflating a tire, what pressure does this produce on the air in the pump?
Answer:
The pressure produced on the air in the pump is 37.209 pounds per square inch.
Explanation:
By definition, the pressure is the force exerted on the piston divided by its area. Given that, force is distributed uniformly on the piston area, the formula to determine the pressure is:
[tex]p = \frac{F}{A}[/tex]
Where:
[tex]p[/tex] - Pressure, measured in pounds per square inch.
[tex]F[/tex] - Force exerted on the piston, measured in pounds.
[tex]A[/tex] - Piston area, measured in square inches.
If [tex]F = 16\,lb[/tex] and [tex]A = 0.43\,in^{2}[/tex], the pressure produced on the air in the pump is:
[tex]p = \frac{16\,lb}{0.43\,in^{2}}[/tex]
[tex]p = 37.209\,psi[/tex]
The pressure produced on the air in the pump is 37.209 pounds per square inch.
which statement did Ernest Rutherford make about atoms?
Answer:
Option A
Explanation:
Ernest Rutherford concluded that the atom has a small, dense center which constitutes the mass of the whole atom. He called it a "Nucleus". He also said that most of the space in the atom is empty.
What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)
Answer:
ΔU = 2.25 x 10⁸ J
t = 104.17 s
Explanation:
The change in internal energy of the engine can be given by the following formula:
ΔU = (Mass of Gasoline)(Energy Content of Gasoline)
ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)
ΔU = 2.25 x 10⁸ J
Now, for the time of operation, we use the following formula of power.
P = W/t = ΔU/t
t = ΔU/P
where,
t = time of operation = ?
ΔU = Change in internal energy = 2.25 x 10⁸ J
P = Power of motor = 600 W
Therefore,
t = (2.25 x 10⁸ J)/(600 W)
t = (375000 s)(1 h/3600 s)
t = 104.17 s
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.
What are Velocity and Acceleration?The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.
There are several types of velocity :
Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform VelocityThe pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.
According to the question, the given values are :
Time, t = 3.50 sec
Initial Velocity, u = 0 m/s
Use equation of motion :
v = u+at
v = 0+ g sin 18 × 3.5
v = 10.6 m/s.
So, the final velocity will be 10.6 m/s.
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According to Newton, when the distance between two interacting objects doubles, the gravitational force is
Answer:
1/4 of its original value
Explanation:
Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e
F ∝ [tex]\frac{M_1 M_2}{r^2}[/tex] ------------(i)
From the equation above, it can be deduced that;
F ∝ [tex]\frac{1}{r^2}[/tex]
=> F = G [tex]\frac{1}{r^2}[/tex] -----------(ii)
Where;
G = constant of proportionality called the gravitational constant
Equation (ii) can be re-written as
Fr² = G
=> F₁r₁² = F₂r₂² -----------(iii)
Where;
F₁ and r₁ are the initial values of the force and distance respectively
F₂ and r₂ are the final values of the force and distance respectively
From the question, if the distance doubles i.e;
r₂ = 2r₁,
Then the final value of the gravitational force F₂ is calculated as follows;
Substitute the value of r₂ = 2r₁ into equation (iii) as follows;
F₁r₁² = F₂(2r₁)²
F₁r₁² = 4F₂r₁² [Divide through by r₁²]
F₁ = 4F₂ [Make F₂ subject of the formula]
F₂ = F₁ / 4 [Re-write this]
F₂ = [tex]\frac{1}{4} F_1[/tex]
Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.
What is Dark Matter?
Explanation:
dark matter is a form of matter thought to account for approximately 85% of the matter in the universe and about a quarter of its total mass–energy density or about 2.241×10⁻²⁷ kg/m³. Its presence is implied in a variety of astrophysical observations, including gravitational effects that cannot be explained by accepted theories of gravity unless more matter is present than can be seen. For this reason, most experts think that dark matter is abundant in the universe and that it has had a strong influence on its structure and evolution. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect.
Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.
Answer:
Objects act differently in a gravity field than in an accelerating reference frame.
Explanation:
The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.
The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.
Answer:
A - The orbital path of mercury around the sun has changed.
Explanation:
got right on edg.
Gravitational potential energy is greatest at the highest point of a roller coaster and least at the lowest point.
Answer:
because gravitational potential enegry is directly proportional to the height so more the height more the gravitational potential enegry. therefore gravitational potential enegry is greatest at high point than lower points.
What accurately depicts the change in average kinetic energy of the particles undergoes in matter as the temperature of the sample is decreased?
Answer:
As a sample of matter is continually cooled, the average kinetic energy of its particles decreases. Eventually, one would expect the particles to stop moving completely. Absolute zero is the temperature at which the motion of particles theoretically ceases.
Explanation:
A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the mass (in kg) of the particle.
Answer:
[tex]v=21.36\,\,\frac{m}{s}\\[/tex]
[tex]m=1.2357\,\,kg[/tex]
Explanation:
Recall the formula for linear momentum (p):
[tex]p = m\,v[/tex] which in our case equals 26.4 kg m/s
and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:
[tex]K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}[/tex]
Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:
[tex]K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg[/tex]
Now by knowing the particle's mass, we use the momentum formula to find its speed:
[tex]p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}[/tex]
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Question:
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Answer:
S = 5.508 × 10¹¹m
V = 2.62 × 10⁸ m/s
Explanation:
The radius of the orbit of Jupiter, Rj is 43.2 light-minutes
radius of the orbit of Mars, Rm is 12.6 light-minutes
Distance travelled S = (Rj - Rm)
= 43.2 - 12.6 = 30.6 light- minutes
= 30.6 × (3 ×10⁸m/s) × 60 s
= 5.508 × 10¹¹m
time = 35mins = (35 × 60 secs)
= 2100 secs
speed = distance/time
V = 5.508 × 10¹¹m / 2100 s
V = 2.62 × 10⁸ m/s
In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true
Answer:
the energy of the photons is greater than the work function of the zinc oxide.
h f> = Ф
Explanation:
In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.
K_max = h f - Ф
in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.
h f > = Ф
A building is located on earth's equator. As the earth rotates about its axis, which floor of the building has the greatest angular speed?
Answer:
The angular speed of the earth rotation is equal. Therefore
Our angular speed due to Earth’s rotation is same at every point on the earth irrespective of the elevation. So your angular speed due to earth’s rotation on the top floor of the building will be same as it is on the ground floor.
Explanation: