Answer:
The current in the small radius loop must be 0.9677 A
Explanation:
Recall that the formula for the magnetic field at the center of a loop of radius R which runs a current I, is given by:
[tex]B=\mu_0\,\frac{I}{2\,R}[/tex]
therefore for the first loop in the problem, that magnetic field strength is:
[tex]B=\mu_0\,\frac{I}{2\,R} =\mu_0\,\frac{3}{2\,(0.093)} =16.129\,\mu_{0}\,[/tex]
with the direction of the magnetic field towards the plane.
For the second smaller loop of wire, since the current goes counterclockwise, the magnetic field will be pointing coming out of the plane, and will subtract from the othe field. In order to the addition of these two magnetic fields to be zero, the magnitudes of them have to be equal, that is:
[tex]16.129\,\,\mu_{0}=\mu_0\,\frac{I'}{2\,R'} =\mu_{0}\,\frac{I'}{2\,(0.03)} \\I'=16.129\,(2)\,(0.03)=0.9677\,\,Amps[/tex]
An electronic stove is rated 1000w, 240v. explain this statement. please I'm in need
Answer and Explanation:
When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.
Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.
Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:
P = IV -------------(i)
Where;
P = Power rating = 1000W
I = Current used
V = Voltage rating = 240V
Substituting these values into equation (i) gives;
1000 = I x 240
I = [tex]\frac{1000}{240}[/tex] = 4.17A
Therefore, the current used by the stove is 4.17A
To get the resistance R of the stove, we can use the relation;
P = [tex]\frac{V^2}{R}[/tex]
R = [tex]\frac{V^2}{P}[/tex]
R = [tex]\frac{240^2}{1000}[/tex]
R = 57.6Ω
Therefore, the resistance of the stove is 57.6Ω
A parallel–plate capacitor is initially charged by connecting it to a battery. The battery is then disconnected. If the distance between the plates is increased, what happens to the charge on the capacitor and the voltage across it?
a. The charge remains fixed and the voltage decreases.
b. The charge decreases and the voltage remains fixed.
c. The charge remains fixed and the voltage increases.
d. The charge decreases and the voltage increases.
Answer:
t the battery of potential difference V be used to charge the capacitor of capacitance C.
∴ the charge stored in the capacitor q=CV
Now the battery is disconnected, so the the charge of the capacitor becomes constant
i.e q=constant OR CV=constant .............(1)
Capacitance of parallel plate capacitor C=
d
Aϵ
o
So if the distance between the plates is increased, then the capacitance will decrease which is compensated by the increase in voltage across the capacitor according to equation (1).
Also the energy stored in the capacitor E=
2C
q
2
⟹E∝
C
1
(∵q=constant)
Thus energy will increase due to the decrease in capacitance.
Explanation:
Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?
Answer:
Each of the capacitor carries the same charge, Q
Explanation:
When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;
[tex]V_T = V_1 + V_2 + V_3[/tex]
We can also calculate this voltage in terms of capacitance and charge;
[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]
Therefore, each of the capacitor carries the same charge, Q
The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.8 AA , how many turns of wire would you need
Answer:
The number of turns of wire needed is 3536 turns.
Explanation:
Given;
length of the wire, L = 8 cm = 0.08 m
magnetic field on the wire, B = 0.1 T
current in the wire, I = 1.8 A
The magnetic field produced by a solenoid is calculated as;
B = μ₀ n I
where;
n is the number of turns per length = N / L
μ₀ is permeability of free space = 4π x 10⁻⁷ N/A²
[tex]B = \frac{\mu_o N I}{L} \\\\N = \frac{BL}{\mu_o I} \\\\N = \frac{0.1 *0.08}{4\pi*10^{-7} *1.8} \\\\N = 3536.32 \ turns[/tex]
Therefore, the number of turns of wire needed is 3536 turns.
An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.
Answer:
[tex]1.357\times 10^{-12}[/tex]
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]
[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]
= [tex]1.357\times 10^{-12}[/tex]
Therefore for computing the capacitance we simply applied the above formula.
The maximum amount of pulling force a truck can apply when driving on
concrete is 10,560 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
Answer:
Explanation:
Let the weight of the truck be W . reaction force R = W
Maximum frictional force = μ R
= .8 x W
So for movement of truck
Pulling force = frictional force
10560 = .8W
W = 13200 N
weight of heaviest truck required = 13200 N .
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, what condition would he have?
A. Nearsightedness
B. Farsightedness
C. Neither nearsightedness nor farsightedness
Answer:
A. NearsightednessExplanation:
A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.
At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.
Cables supporting a suspension bridge have a linear mass density of 3700 kg/m; the tension is 1.7 x 10 8 N. What would be the transverse wave speed in such a cable?
Answer:
The speed is [tex]v = 214.35 \ m/s[/tex]
Explanation:
From the question we are told that
The linear mass density is [tex]\mu = 3700 \ kg/m[/tex]
The tension is [tex]T = 1.7*10^8 \ N[/tex]
The transverse wave speed is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{1.7 *10^{8}}{3700} }[/tex]
[tex]v = 214.35 \ m/s[/tex]
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.
The medical profession divides the ultraviolet region of the electromagnetic spectrum into three bands: UVA (320-420 nm), UVB (290-320 nm), and UVC (100-290 nm). UVA and UVB promote skin cancer and premature skin aging; UVB also causes sunburn, but helpfully fosters production of vitamin D. Ozone in Earth's atmosphere blocks most of the more dangerous UVC. Find the frequency range associated with UVB radiation.
Answer:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
Explanation:
The frequency of an electromagnetic radiation can be given by the following formula:
υ = c/λ
where,
υ = frequency of electromagnetic wave = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of electromagnetic wave = 290 nm to 320 nm
FOR LOWER LIMIT OF FREQUENCY:
λ = 320 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)
υ = 9.375 x 10¹⁴ Hz
FOR UPPER LIMIT OF FREQUENCY:
λ = 290 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)
υ = 10.34 x 10¹⁴ Hz
Therefore, the frequency range for UVB radiations is:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
calculate the upthrust aciting on a body if its
true weight is 550 N and apparent weight
lis 490 N
Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s. m/s
Answer:
a)
6t i + t² j m
b)
6 i + 2 t j m/s
c)
x = 24 m and y = 16 m
d)
10 m/s
Explanation:
Explanation is given in the attached document.
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)
Answer:
Power=50.17dioptre
Power=50.17D
Explanation:
P=1/f = 1/d₀ + 1/d₁
Where d₀ = the eye's lens and the object distance= 5.70m=
d₁= the eye's lens and the image distance= 0.02m
f= focal length of the lense of the eye
We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m
Therefore, we can calculate the power using above formula
P= 1/5.70 + 1/0.02
Power=50.17dioptre
Therefore, the power the eye's is using to see the object from distance is 5.70D
An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the car travel before stopping
Answer:
344.68 m
Explanation:
The computation of the far does the car travel before stopping is
Data provided in the question
Force = F = 8,868 N
mass = m = 2,500 kg
So,
accleration = a is
[tex]= \frac{-F}{m}\\\\\= \frac{8868}{2500}[/tex]
a = -3.54 m/s^2
The initial speed = u = 49.4 m / s
final speed = v = 0
Based on the above information
Now applying the following formula
v^ 2- u^ 2= 2aS
Therefore
[tex]S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}[/tex]
= 344.68 m
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The moment about the origin O of the force F is Mo
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t
Answer:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Explanation:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as originating from a point on the near side of the mirror.
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved mirror than the centre of curvature.
During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and Fz of the net force on the airplane.
Answer:
[tex]F_x[/tex] = -1.5t
[tex]F_y[/tex] = 0.25
[tex]F_{z}[/tex] = 0
Explanation:
Given equation;
p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.
From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e
∑F = [tex]\frac{dp}{dt}[/tex] -----------(i)
Substitute the equation of p into equation (i) and differentiate with respect to t as follows;
∑F = [tex]\frac{dp}{dt}[/tex] = [tex]\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}[/tex]
∑F = [tex]\frac{dp}{dt}[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex]
∑F = [tex][-1.5t + 0]i + 0.25j[/tex]
But
∑F = [tex]F_xi + F_yj + F_zk[/tex]
Where;
[tex]F_x, F_y, F_z[/tex] are the components of the net force in the x, y and z direction respectively.
=> [tex]F_xi + F_yj + F_zk[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex] = [tex]-1.5ti + 0.25j[/tex]
=> [tex]F_x[/tex] = -1.5t
=> [tex]F_y[/tex] = 0.25
=> [tex]F_{z}[/tex] = 0
A stellar object is emitting radiation at 3.55 mm. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?
Answer:
E = 6.45 x 10⁻¹¹ J
Explanation:
First we need to find total number of photons detected in 1 hour. Therefore,
No. of Photons = n = (3.2 x 10⁸ photons/s)(1 h)(3600 s/1 h)
n = 11.52 x 10¹¹ photons
Now, the energy of these photons can be given by the formula:
E = nhc/λ
where,
E = Total Energy of the Photons = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of radiation = 3.55 mm = 3.55 x 10⁻³ m
Therefore,
E = (11.52 x 10¹¹)(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.55 x 10⁻³ m)
E = 6.45 x 10⁻¹¹ J
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and 300 K, respectively. The heat transfer from the air during the isothermal compression is 80 kJ. At the end of the isothermal compression, the volume is 0.2 m3. Determine the volume at the beginning of the isothermal compression, in m3. Assume the ideal gas model for air and neglect kinetic and potential energy effects.
Answer:
Explanation:
Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .
So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .
Work done on gas at isothermal compression is equal to heat transfer .
work done on gas = 80 x 10³ J
work done on gas = n RT ln v₁ / v₂
n is number of moles v₁ and v₂ are initial and final volume
molecular weight of gas = 28.97 g
1.5 kg = 1500 / 28.97 moles
= 51.77 moles
work done on gas = n RT ln v₁ / v₂
Putting the values in the equation above
80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2
ln v₁ / .2 = .62
v₁ / .2 = 1.8589
v₁ = 0.37 m³
One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic field along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]
Therefore, the voltage induced in the coil is 1.25 V.
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
A student is trying to decide what to wear.His bedroom is at 20.0 °C.His skin temperature is 35.0 °C.The area of his exposed skin is 1.50 m².People all over the world have skin that is dark in the infrared,with emissivity about 0.900.Find the net energy transfer from his body by radiation in 10.0 min.
Answer:
vgghcgkxcjpfiffj,ncfzfzujfzxxxoifkc xuzrusdoxTXcxgifdhh
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
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
A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.0 N on the block. When the block is released, it oscillates with a frequency of 1.2 Hz. How far was the block pulled back before being released?
Answer:
Explanation:
For spring
[tex]n=\sqrt{\frac{k}{m} }[/tex]
where n is frequency of oscillation and k is force constant and m is mass
Putting the values
[tex]1.2=\sqrt{\frac{k}{.28} }[/tex]
k = .4032 N/m
F= k x
where F is force , k is force constant and x is extension
Putting the given values
1 = .4032 x
x = 2.48 m
Consider a house at sea level that has 2500 square feet of floor area. What is the total force that the air inside the house exerts upward on the ceiling
Answer:
Total Force that the air inside the house exerts upward on the ceiling is 5.25 × 10⁶ lb
Explanation:
Force = Atmospheric Pressure × Floor Area
Where; Standard Atmospheric Pressure = 2100 lb/ft²
Floor Area = 2500 ft²
Substitute the data
∴ Total Force = 2100 lb/ft² × 2500 ft²
Total Force = 5.25 × 10⁶ lb
