Answers
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
Related Questions
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answers
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
Answers
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998.0 rad/s while it is being read, and then is allowed to come to rest over 0.502 seconds , what is the magnitude of the average angular acceleration of the disk?
Answers
Answer:
1988.05 rad/s^2
Explanation:
The angular speed of the optical disk ω = 998.0 rad/s
the time taken to come to rest t = 0.502 s
The magnitude of the average angular acceleration ∝ = ω/t
∝ = 998.0/0.502 = 1988.05 rad/s^2
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye
Answers
Answer:
30.93 cm
Explanation:
Given that:
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses
The power of the old pair of lens p = 2.25 diopters
The focal point length = 1/p
The focal point length = 1/2.25
The focal point length = 0.444 m
The focal point length = 44.4 cm
The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens
The near point of the person from the glass = 83 cm
Let consider s' to be the image on the same sides of the lens,
∴ s' = -83 cm
We known that:
the focal length of a mirror image 1/f =1/u +1/v
Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.
Then:
1/f = 1/s + 1/s'
1/s = 1/f - 1/s'
1/s = (s' -f)/fs'
s = fs'/(s'-f)
s =( 44.4× -83)/(-83 - 44.4)
s = - 3685.2 / - 127.4
s = 28.93 cm
Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm
= 30.93 cm
The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related
Answers
Answer:
The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].
Explanation:
Net force on the hoop is given by;
[tex]F_{net} = ma[/tex]
where;
a is linear acceleration
m is the mass
Net torque on the hoop is given by;
[tex]\tau_{net} =I\alpha[/tex]
where;
I is moment of inertia
α is the angular acceleration
But, τ = Fr
[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]
[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]
let the angular acceleration of the smaller hoop = α₁
let the radius of the smaller hoop = r₁
then, the radius of the larger loop, r₂ = 2r₁
let the angular acceleration of the larger hoop = α₂
[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]
Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]
A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N
Answers
Answer:
The correct answer is 500 N
Explanation:
This is an exercise in Newton's third law or law of action and reaction
The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.
Which force we call action does not matter, the analysis and conclusions are the same
The correct answer is 500N
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answers
Answer:I don’t know
Explanation:
Using a conventional two-slit apparatus with light of wavelength 605 nm, 34 bright fringes per centimeter are observed on a screen 3.1 m away. What is the slit separation
Answers
Answer:
d = 6.38 x 10⁻³ m = 6.38 mm
Explanation:
Since, the no. of bright fringes is 34 in a centimeter, therefore, the fringe spacing must be equal to:
Fringe Spacing = Δx = 1 cm/34
Δx = 0.0294 cm = 2.94 x 10⁻⁴ m
But, the formula for fringe spacing in a double slit experiment is:
Δx = λL/d
where,
λ = wavelength of light = 605 nm = 6.05 x 10⁻⁷ m
L = Distance between screen and slits = 3.1 m
d = slit separation = ?
Therefore,
2.94 x 10⁻⁴ m = (6.05 x 10⁻⁷ m)(3.1 m)/d
d = (18.755 x 10⁻⁷ m²)/(2.94 x 10⁻⁴ m)
d = 6.38 x 10⁻³ m = 6.38 mm
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Answers
Answer:
b) True. potencial diferencie does not depend on orientation
Explanation:
In this exercise we are asked to show which statements are true.
The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.
It does not refer to the height of the system.
We can now review the claims
a) False. Potential not to be refers to height
b) True. Does not depend on orientation
c) False The potential does not refer to the altitude but to the Earth's charge
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answers
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?
Answers
Answer:
A. 4.82 cm
B. 24.66 cm
Explanation:
The depth of water = 19.6 cm
Distance of fish = 6.40 cm
Index of refraction of water = 1.33
(A). Now use the below formula to compute the apparent depth.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]
(B). the depth of the fish in the mirror.
[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]
Now find the depth of reflection of the fish in the bottom of the tank.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]
550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answers
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answers
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answers
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or occurring at constant pressure. Rank those three processes in order of the work required of an external agent, least to greatest. A : adiabatic, isothermal, constant pressure B : constant pressure, isothermal, adiabatic C : adiabatic, constant pressure, isothermal D : isothermal, adiabatic, constant pressure E : constant pressure, adiabatic, isothermal
Answers
Answer:
B. constant pressure, isothermal, adiabatic
Explanation:
A quantity of an ideal gas is compressed to half its initial volume.
The process may be adiabatic, isothermal or occurring at constant pressure.
Adiabatic-constant heat
Constant pressure or isobaric
Isothermal or constant temperature
An external agent is a system that does work on a system or a machine.
This external agent applies force , or changes the state of the body it is acting on.
In order of the work required of an external agent, least to greatest
The following processes will be arranged.
constant pressure, isothermal, adiabatic
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low- mass sail and the energy and momentum of sunlight for propulsion.
Should the sail be absorbing or reflective? Why?
a. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is smaller than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
b. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is larger than for reflective sail, therefore the radiation pressure is larger for the absorbing sail
c. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is smaller than for reflective sail, therefore the radiation pressure is larger for the absorbing sail.
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
Answers
Answer:
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
We've seen that for thermal radiation, the energy is of the form AVT4, where A is a universal constant, V is volume, and T is temperature. 1) The heat capacity CV also is proportional to a power of T, Tx. What is x
Answers
Answer:
this raise the temperature is x = 3
Explanation:
Heat capacity is the relationship between heat and temperature change
C = Q / ΔT
if the heat in the system is given by the change in energy and we carry this differential formulas
[tex]c_{v}[/tex] = dE / dT
In this problem we are told that the energy of thermal radiation is
E = A V T⁴
Let's look for the specific heat
c_{v} = AV 4 T³
the power to which this raise the temperature is x = 3
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Answers
Complete Question
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex] in SI units.
Answer:
The value is [tex]f = 1.98918*10^{5}\ Hz[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]
This above equation can be modeled as
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]
So
[tex]w = \frac{10^7}{8}[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{w}{2 \pi}[/tex]
=> [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]
=> [tex]f = 1.98918*10^{5}\ Hz[/tex]
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answers
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon
Answers
Answer:
The new volume will be 0.918 x 10^5 L
Explanation:
initial volume [tex]V_{1}[/tex] = 1.21 x 10^5 L
Initial temperature [tex]T_{1}[/tex] = 265 K
Final volume [tex]V_{2}[/tex] = ?
Final temperature [tex]T_{2}[/tex] = 201 K
Th gas is an ideal gas.
For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant
substituting value, we have
(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201
[tex]V_{2}[/tex] = 24321000/265 = 91777.4 L
= 0.918 x 10^5 L
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answers
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answers
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answers
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answers
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
Two protons moving with same speed in same direction repel each other but what about two protons moving with different speed in the same direction?
Answers
Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.
In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force
Explanation:
Explanation:
Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answers
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answers
Answer:
C $0.75 my friend I wish it is right answer
An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .
Required:
a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
b. What would be the angular position of the second-order, two-slit, interference maxima in this case?
Answers
Answer:
a. 0.058°
b. 0.117°
Explanation:
a. The angular position of the first-order is:
[tex] d*sin(\theta) = m\lambda [/tex]
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]
Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.
b. The angular position of the second-order is:
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]
Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.
I hope it helps you!
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of the circular motion of the alpha particles is closest to: A. 0.25 μs B. 0.16 μs C. 0.49 μs D. 0.40 μs E. 0.33 μs
Answers
Answer:
Option B: T ≈ 0.16 μs
Explanation:
We are given;
Mass; m = 6.68 × 10^(-27) kg
Magnetic field;B = 0.80 T
Charge;q = 2e
Now, e is the charge on an electron and it has a value of 1.6 × 10^(-19) C
So, q = 2 × 1.6 × 10^(-19)
q = 3.2 × 10^(-19) C
The period of the circular motion of the alpha particles moving along a in the presence of the magnetic field is given by;
T = 2πm/qB
Where ;
m, q and B are as stated earlier.
Plugging in the relevant values, we have;
T = (2π × 6.68 × 10^(-27))/(3.2 × 10^(-19) × 0.8)
T = 0.16395 × 10^(-6) s
This can also be written as;
T ≈ 0.16 μs
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answers
Answer:
1.125×10⁻⁹ J
Explanation:
Applying,
E = 1/2CV²................... Equation 1
Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.
Given; C = 1.0 nF, = 1.0×10⁻⁹ F, V = 1.5 V
Substitute into equation 1
E = 1/2(1.0×10⁻⁹×1.5²)
E = 1.125×10⁻⁹ J
Hence the energy stored by the capacitor is 1.125×10⁻⁹ J
