Answer:
the correct answer is a
Explanation:
The torque is
τ = F x r
where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body
τ = mg r
in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero
the correct answer is a
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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a bullet is dropped from the same height when another bullet is fired horizontally they will hit the ground
Answer:
simultaneously
Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.
As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal
Answer:
(A) Gravity is you're answer.
Explanation:
When an object or human is falling at an increased rate, The force of gravity is taking place.
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
[tex]F=133N[/tex]
Explanation:
From the question we are told that:
Length [tex]l=3.0m[/tex]
Mass [tex]m=24kg[/tex]
Distance from Tip [tex]d=35cm[/tex]
Generally, the equation for Torque Balance is mathematically given by
[tex]mg(l/2)=F(l-d)[/tex]
[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]
Therefore
[tex]F=133N[/tex]
Your little sister (mass 25 kg) is sitting in her little red wagon (mass
8.5 kg) at rest. You begin pulling her forward, accelerating her with a
constant force for 2.35 s to a speed of 1.8 m/s. Calculate the impulse
you imparted to the wagon and its passenger.
Answer:
p = 60.6N*s
Explanation:
v_f = v_0+a*t
a = (v_f-v_0)/t
a = (1.8m/s)/2.35s
a = 0.77m/s²
F = m*a
F = (25kg+8.5kg)*0.77m/s²
F = 25.8N
^p = F*t
p = 25.8N*2.35s
p = 60.6N*s
Air is compressed polytropically from 150 kPa, 5 meter cube to 800 kPa. The polytropic exponent for the process is 1.28. Determine the work per unit mass of air required for the process in kilojoules
a) 1184
b) -1184
c) 678
d) -678
Answer:
wegkwe fhkrbhefdb
Explanation:B
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
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When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
what is the major difference between the natural frequency and the damped frequency of oscillation.
Answer:
This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s at 60° above the horizontal. Calculate (a) the maximum height, (b) the time required to reach its highest point, (c) the total time of flight, (d) the components of its velocity just before striking the ground, and (e) the horizontal distance traveled from the base of the cliff.
a) y(max) = 337.76 m
b) t₁ = 5.30 s the time for y maximum
c)t₂ = 13.60 s time for y = 0 time when the fly finish
d) vₓ = 30 m/s vy = - 81.32 m/s
e)x = 408 m
Equations for projectile motion:
v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )
v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s
a) Maximum height:
The following equation describes the motion in y coordinates
y = y₀ + v₀y*t - (1/2)*g*t² (1)
To find h(max), we need to calculate t₁ ( time for h maximum)
we take derivative on both sides of the equation
dy/dt = v₀y - g*t
dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g
v₀y = 60*sin60° = 60*√3/2 = 30*√3
g = 9.8 m/s²
t₁ = 5.30 s the time for y maximum
And y maximum is obtained from the substitution of t₁ in equation (1)
y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²
y (max) = 200 + 275.40 - 137.64
y(max) = 337.76 m
Total time of flying (t₂) is when coordinate y = 0
y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²
0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0
The above equation is a second-degree equation, solving for t₂
t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8
t = [51.96 ±√2700 + 3920]/9.8
t = [51.96 ± 81.36]/9.8
t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)
t₂ = 13.60 s time for y = 0 time when the fly finish
The components of the velocity just before striking the ground are:
vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant
vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)
vy = 51.96 - 133.28 vy = - 81.32 m/s
The sign minus means that vy change direction
Finally the horizontal distance is:
x = vₓ * t
x = 30 * 13.60 m
x = 408 m
A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration
Answer:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
Amplitude = 24 cm
May someone help...please. Pretty please...
If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height? Assume that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same.
f(short)/f(avg)=?
We have that the ratio of his walking pace to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
given the assumption and the calculation given below
From the question we are told that:
Consider a person 18\% shorter than average
Let average height of a person be [tex]10m[/tex]
Therefore
The height of an [tex]18\%[/tex] shorter man is mathematically given as
H=10*0.18
H=8.2m
Generally, the equation for velocity is mathematically given by
[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]
Where we have the Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same
Therefore
[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]
[tex]\frac{V_1}{V_2}={82}{100}[/tex]
[tex]\frac{V_2}{V_1}=1.10[/tex]
In conclusion
The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
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There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
14. What's one of the two requirements electric current?
A. There must be an electric potential between two bodies
B. There must be no valence electrons that make their element unstable
C. There must be a carbon element present in the electric current
D. There must be a magnetic force between two bodies
Marko
One of the two requirements of electric current is there must be an electric potential between two bodies
For electric current to flow, there must be an electric potential between two bodies.
This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.
The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.
This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.
So, one of the two requirements of electric current is there must be an electric potential between two bodies.
So, the answer is A
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Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
What is utilization of energy
Explanation:
Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector
what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop
Answer:
The initial velocity is 1.27 m/s.
Explanation:
distance, s = 1.8 m
acceleration, a = - 0.45 m/s^2
final velocity, v = 0
let the initial velocity is u.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]
We have that the Initial velocity is mathematically given as
u=1.27m/s
Maximum possible velocity
Question Parameters:
a slight push toward an end stop 1.80 meters away
he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2
Generally the equation for the third equation of motion is mathematically given as
Vf^2 = Vi^2 + 2ad
Therefore
0=u^2+0.45*1.8
u=1.27m/s
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Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát
Explanation:
mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
