Answer:
The magnetic field strength : 0.372 T
Explanation:
The equation of the induced emf is given by the following equation,
( Equation 1 ) emf = - N ( ΔФ / Δt ) - where N = number of turns of the coil, ΔФ = change in the magnetic flux, and Δt = change in time
The equation for the magnetic flux is given by,
( Equation 2 ) Ф = BA( cos( θ ) ) - where B = magnetic field, A = area, and θ = the angle between the normal and the magnetic field
The area of the circular coil is a constant, as well as the magnetic field. Therefore the change in the magnetic flux is due to the angle between the normal and the magnetic field. Therefore you can expect the equation for the change in magnetic flux to be the same as the magnetic flux, but only that there must be a change in θ.
( Equation 3 ) ΔФ = BA( Δcos( θ ) )
Now as the coil rotates one-fourth of a revolution, θ changes from 0 degrees to 90 degrees. The " change in cos θ " should thus be the following,
Δcos( θ ) = cos( 90 ) - cos( 0 )
= 0 - 1 = - 1
Let's substitute that value in the third equation,
( Substitution of Δcos( θ ) previously, into Equation 3 )
ΔФ = BA( - 1 ) = - BA
Remember the first equation? Well if the change in the magnetic flux = - BA, then through further substitution, the emf should = - N( - BA ) / Δt. In other words,
emf = - N( - BA ) / Δt,
emf = NBA / Δt,
B = ( emf )Δt / NA
Now that we have B, the magnetic field strength, isolated, let's solve for the area of the circular coil and substitute all known values into this equation.
Area ( A ) = πr²,
= π( 0.275 )² = 0.2376 m²,
B = ( 10,000 V )( 4.42 [tex]*[/tex] 10⁻³ s ) / ( 500 )( 0.2376 m² ) = ( About ) 0.372 T
The magnetic field strength : 0.372 T
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.
Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.
Explanation:
It is given that,
The separation between two parallel wires, r = 11 cm = 0.11 m
Current in wire 1, [tex]q_1=54\ A[/tex]
Current in wire 2, [tex]q_2=45\ A[/tex]
Length of wires, l = 4.3 m
We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]
So, the magnetic force on a 4.3 m length of the wire on both of currents is F=0.0189 N.
You throw a stone vertically upward with a speed of 26.0 m/s. (a) How fast is it moving when it reaches a height of 15.0 m? (b) How much time is required to reach this height when it's falling down? a. 19.5 m/s , b. 4.51 s a. 17.9 m/s , b. 0.620 s a. 19.5 m/s , b. 0.800 s a. 17.9 m/s , b. 4.28 s a. 380 m/s , b. 8 s
Answer:
ok well
Explanation:
teghe
Answer:
v = 19.5 m/s
t = 4.51 s
Explanation:
a)
given:
height is 15m from the ground
initial velocity Vi = 26 m/s
acceleration a or g = 9.81 m/s²
formula: Vf² = Vi² + 2aΔy
26² = Vi² + 2 (9.81) 15
Vi = 19.5 m/s
now you can calculate the time by using the equations below:
Δy = 1/2 (Vi + Vf) t
Vf = Vi + a t
Δy = Vi t + 1/2 a t
time must be 4.51 s
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.
Required:
a. How much work is done by the tension?
b. How much work is done by gravity?
The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
The given parameters;
mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 NThe work done by the tension is calculated as follows;
W = Fd
W = 15 x 0.15
W = 2.25 J
The work done by gravity is calculated as;
W = (25 x 9.8) x 0.15
W = 36.75 J
Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
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An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part of the process she would follow?
A. plot an H-R diagram for the stars in the cluster
B. count the number of M type stars in the cluster
C. measure the Doppler shift of a number of the stars in the cluster
D. search for planets like Jupiter around the stars in the center of the cluster
E. search for x-rays coming from the center of the cluster
Answer:
A. plot an H-R diagram for the stars in the cluster.
Explanation:
A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.
The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?
a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.
Answer:
the greatest intensity is obtained from c
Explanation:
An electromagnetic wave stagnant by the expression
E = E₀ sin (kx -wt)
when two waves meet their electric fields add up
E_total = E₁ + E₂
the intensity is
I = E_total . E_total
I = E₁² + E₂² + 2E₁ E₂ cos θ
where θ is the phase angle between the two rays
Let's examine the two waves
in this case E₁ = E₂ = E₀
I = Eo2 + Eo2 + 2 E₀ E₀ coasts
I = E₀² (2 + 2 cos θ )
I = 2 I₀ (1 + cos θ )
let's apply this expression to different cases
a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀
b) cos π = -1 this implies that I_total = 0
c) the cosine is 1,
I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ
I = E₀² (5 +4 cos θ)
I = E₀² 9
I = 9 Io
d) in this case the cos pi = -1
I = E₀² (5 -4)
I = I₀
e) we rewrite the equation
I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ
I = Eo2 (10 + 6 cos θ)
cos π = -1
I = E₀² (10-6)
I = 4 I₀
the greatest intensity is obtained from c
The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.
What is an amplitude?An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.
In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.
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The angle between the axes of two polarizing filters is 41.0°. By how much does the second filter reduce the intensity of the light coming through the first?
Answer:
The amount by which the second filter reduces the intensity of light emerging from the first filter is
z = 0.60
Explanation:
From the question we are told that
The angle between the axes is [tex]\theta = 41^o[/tex]
The intensity of polarized light that emerges from the second filter is mathematically represented as
[tex]I= I_o cos^2 \theta[/tex]
Where [tex]I_o[/tex] is the intensity of light emerging from the first filter
[tex]I = I_o [cos(41.0)]^2[/tex]
[tex]I =0.60 I_o[/tex]
This means that the second filter reduced the intensity by z = 0.60
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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change in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius
Answer:
4519.60 J/KExplanation:
Change in entropy is expressed as ΔS = ΔQ/T where;
ΔQ is the total heat change during conversion of ice to water.
T is the room temperature
First we need to calculate the total change in heat using the conversion formulae;
ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)
m is the mass of ice/water = 900g = 0.9kg
L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg
c is the specific heat capacity of water = 4200J/kgK
Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K
Substituting the given values into ΔQ;
ΔQ = 0.9(333000)+0.9(4200)(283)
ΔQ = 299700 + 1069740
ΔQ = 1,369,440 Joules
Since Change in entropy ΔS = ΔQ/T
ΔS = 1,369,440/30+273
ΔS = 1,369,440/303
ΔS = 4519.60 J/K
Hence, the change in entropy of the universe is 4519.60 J/K
Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy
Answer:
D. Light is a form of energy
Explanation:
A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:
5.19 x 10³Hz
Explanation:The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;
[tex]X_C = \frac{1}{2\pi fC }[/tex]
Where;
f = frequency of the signal through the capacitor
C = capacitance of the capacitor.
Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;
V = I x [tex]X_{C}[/tex] [Substitute the value of
=> V = I x [tex]\frac{1}{2\pi fC}[/tex] [Make f the subject of the formula]
=> f = [tex]\frac{I}{2\pi VC}[/tex] ---------------------(i)
From the question;
I = 3.33mA = 0.00333A
C = 8.50nF = 8.50 x 10⁻⁹F
V = 12.0V
Substitute these values into equation (i) as follows;
f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex] [Taking [tex]\pi[/tex] = 3.142]
f = 5.19 x 10³Hz
Therefore, the frequency is closest to f = 5.19 x 10³Hz
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.
Answer:
Explanation:
The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.
In compound microscopes, the distance between the two lenses is expressed as L = v0+ue
v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.
Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.
Given L = 18.0cm
Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm
1/f0 = 1/u0+1/v0
f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.
1/v0 = 1/7.4-1/12
1/v0 = 0.1351-0.0833
1/v0 = 0.0518
v0 = 1/0.2184
v0 = 19.31cm
Note that v0 = ue = 19.31cm
To get ve, we will use the lens formula 1/fe = 1/ue+1/ve
1/ve = 1/fe-1/ue
Given ue = 19.31cm and fe = -7.00cm
1/ve = -1/7.0-1/19.31
1/ve = -0.1429-0.0518
1/ve = -0.1947
ve = 1/-0.1947
ve = -5.14cm
Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens
b) Magnification of the final image M = ve/ue
M = 5.14/19.31
M = 0.27
Magnification of the final image is 0.27
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
Express your answer numerically in kilowatts to two significant figures.
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Express the total force numerically, to two significant figures, in units of newtons.
Answer:
1) 14 kW
2) 4.67 x 10^-5 N
Explanation:
Area of solar panel = 10 m^2
Intensity of sun's radiation incident on earth = 1.4 kW/m^2
Solar power absorbed = ?
We know that the intensity of radiation on a given area is
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where I is the intensity of the radiation
P is the power absorbed due to this intensity on a given area
A is the area on which this radiation is incident
From the equation, we have
P = IA
P = 1.4 kW/m^2 x 10 m^2 = 14 kW
b) For a perfect absorbing surface, the radiation pressure is given as
p = I/c
where p is the radiation pressure
I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa
we know that Force = pressure x area
therefore force on the solar panels is
F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
Find out more information about Force here:
https://brainly.com/question/25239010
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2
Answer:
Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram
Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely
used integrated circuits for creating clock pulses is called a 555 timer. shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?
ANSWER : R1 = 144.3Ω, R2 = 72.2 Ω
Explanation:
Frequency = 10 MHz
Time period = 1 / F = 0.1 u s
Duty cycle = 75% = 0.75
Duty cycle can be represented as : Ton / T
Also: Ton = Th = 0.75 * 0.1 u s = 75 n s
TL = T - Th = 100 ns - 75 n s = 25 n s
To find the value of R2 we use the equation for time spent in the low (0 V) state
TL = R2*C*ln(2)
hence R2 = TL / ( C * In 2 )
c = 500 pF
Hence R2 = 25 / ( 500 pF * 0.693 ) = 72.2 Ω
To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,
Th = (R1 + R2)*C*ln(2)
from the equation make R1 the subject of the formula
R1 = (Th - ( R2 * C * In2 )) / (C * In 2)
R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )
R1 = ( 75 ns - ( 25 ns ) / 500 pf * 0.693
= 144.3Ω
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s