Answer:
Visible light intensity at the surface of the bulb (I) = 331 W/m²
Explanation:
Given:
Energy = 75 W
Radius = 6 /2 = 3 cm = 3 × 10⁻² m
Energy goes to visible light = 5% = 0.05
Find:
Visible light intensity at the surface of the bulb (I)
Computation:
Visible light intensity at the surface of the bulb (I) = P / 4A
Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²
Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)
Visible light intensity at the surface of the bulb (I) = 331 W/m²
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in the primary if the secondary load resistance is 250 W?
Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, [tex]N_P[/tex] = 50 turns
number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns
the secondary load resistance, [tex]R_S[/tex] = 250 Ω
Determine the turns ratio;
[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]
Now, determine the reflected resistance in the primary winding;
[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]
Therefore, the reflected resistance in the primary winding is 6250 Ω
Give an example of a fad diet that is not healthy and one that is healthy. Explain how you know the difference.
Answer:
Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.
Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.
Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.
Answer:
A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods.
Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead.
Explanation:
Got a 100
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the lens.
Part A. How far must the lens be from the photocells?
s = cm
Part B. Is the image on the photocells erect or inverted? Real or virtual?
a. The image is erect and real.
b. The image is inverted and real.
c. The image is erect and virtual.
d. The image is inverted and virtual.
Part C. How tall is the image on the photocells?
|h?| = cm
Part D. A SLR digital camera often has pixels measuring 8.00?m
Answer:
a. 7.62cm
b. Real and inverted
c. 2.76 cm
d. 3450
Explanation:
We proceed as follows;
a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;
1/f = 1/p + 1/q
making q the subject of the formula;
q = pf/p-f
From the question;
p = 4.70m
f = 7.5cm = 0.075m
Substituting these, we have ;
q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm
b. The image is real and inverted since the image distance is positive
c. We want to calculate how tall the image is
Mathematically;
h1 = (q/p)h0
h1 = (7.62/4.70)* 1.7
h1 = 2.76 cm
d. We want to calculate the number of pixels that fit into this image
Mathematically:
n = h1/8 micro meter
n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a force F? =88i^?23j^ to the end of the wrench. What is the torque on the bolt?
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
A rigid uniform bar of length L and mass m is suspended by a massless wire AC and a rigid massless link BC. Determine the tension in BC immediately after AC breaks.
Answer:
hello the needed diagram is missing attached below is the diagram and the detailed solution
The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]
Explanation:
ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM
Ma = mg - T/ [tex]\sqrt{2}[/tex] equation 1
Ma = 3T / [tex]\sqrt{2}[/tex] equation 2
equate both equations to determine the tension on BC
48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?
Answer:
the interation blood veins
Explanation:
The valid digits in a measurement are called _____ digits. Question 10 options: insignificant significant uncertain non-zero
Answer:
Significant
Explanation:
Valid digits in measurements are called significant digits, or also called significant figures.
These significant digits allow data and measurements to be more accurate and exact.
Answer:
Significant digits
Explanation
Took the test got it right
A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
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A car travels down the road for 535 m in 17.3 s. What is the velocity of the car in m/s and in km/h?
Answer:
30.92m/sExplanation:
[tex]Distance = 535m\\Time = 17.3s\\\\Velocity = \frac{Distane}{Time} \\\\V = \frac{535m}{17.3s} \\\\Velocity = 30.92m/s[/tex]
[tex]Distance = 535m\\\\535m \:to \: km=0.535km\\\\Time = 17.3s\\\\17.3s = 0.004805556hours\\\\Velocity = \frac{Distance}{Time}\\\\ V= \frac{0.535}{0.004805556} \\\\ V=111.329469472\\\\=111.33km/h[/tex]
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
a. no magnetic field exists in that region of space.
b. the particle must be moving parallel to the magnetic field.
c. the particle is moving at right angles to the magnetic field.
d. either no magnetic field exists or the particle is moving parallel to the magnetic field.
e. either no magnetic field exists or the particle is moving perpendicular to the magnetic field.
Answer:
b. the particle must be moving parallel to the magnetic field.
Explanation:
The magnetic force on a moving charged particle is given by;
F = qvBsinθ
where;
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
θ is the angle between the magnetic field and velocity of the moving particle.
When is the charge is stationary the magnetic force on the charge is zero.
Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.
Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.
b. the particle must be moving parallel to the magnetic field.
A uniform narrow tube 1.90 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 294 Hz.(a) What is the fundamental frequency?_____Hz(b) What is the speed of sound in the gas in the tube?________ m/s
Answer:
a)14Hz
b)26.6m/s
Explanation:
a)we were given
the first harmonics frequencies as 280 Hz
The second harmonic frequency as 294 Hz.
The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then
the fundamental frequency=(294 - 280 = 10 Hz)
= 14Hz
b) We know the frequency and the wavelength of the sound wave (
We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as
And fundamental frequency= 14Hz, and distance of 1.90 m then
v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s
Therefore, the speed of sound in the gas in the tubes is 26.6m/s
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?
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 child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
[tex]x - x_o = u_xt[/tex]
[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex] ---- (1)
the equation of the motion y is :
[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]
[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]
[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]
[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]
[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]
[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]
By using the quadratic formula, we have;
[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]
where;
a = 4.9, b = -5.14 c = 1
[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]
[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
[tex]\mathtt{x = u_x(0.258)}[/tex]
[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]
[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]
[tex]\mathbf{x = 1.581 \ m}[/tex]
Thus, the child is 1.581 m far from the fence
What is the mass of a rectangular block of
density 2.5 ×10³ k gm³that measures 10cm by 5 cm by 4 cm?
A. 0.002 kg
B. 0.080 kg
C. 0.200 kg
D. 0.500 kg
E. 1.000 kg
Answer:
Option (D) : 0.5 kg
Explanation:
[tex]mass = density \times volume[/tex]
[tex]mass = {2500} \times 0.1 \times 0.05 \times 0.04[/tex]
Mass of block = 0.5 kg
the mass of a rectangular block of density 2.5 ×10³ k gm³ that measures 10cm by 5 cm by 4 cm is 0.5 kg.
What is density ?Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water)
The Volume of the block is,
V = LBD, where L = length, B = breadth , D = depth of the block.
V = 10 × 5 × 4 = 200 cm³
Density of Block = 2.5 ×10³ kg/m³
Density = Mass / Volume
2.5 ×10³ kg/m³ = Mass / 200 cm³
2.5 ×10³ kg/m³ × 200 cm³ = Mass
2.5 ×10³ kg/m³ × 0.2 × 10⁻³ m³ = Mass
Mass = 0.5 kg
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Consider a hydraulic lift that uses an input piston with an area of 0.5m2. An input force of 15N is exerted on this piston. If the output piston has an area of 3.5m? What is the output force?
Answer:
The output force of the piston is 105 N.
Explanation:
Given;
the area of the input piston, A₁ = 0.5 m²
the input force of the piston, F₁ = 15 N
the area of the output piston, A₀ = 3.5 m²
the output force of the piston, F₀ = ?
The pressure of the hydraulic lift is given by;
[tex]P = \frac{F}{A}[/tex]
where;
P is the hydraulic pressure
F is the piston force
A is the area of the piston
[tex]P = \frac{F}{A} \\\\\frac{F_o}{A_o} = \frac{F_i}{A_i} \\\\F_o = \frac{F_iA_o}{A_i} \\\\F_o = \frac{15*3.5}{0.5} \\\\F_o = 105 \ N[/tex]
Therefore, the output force of the piston is 105 N.
The intensity level 10 m from a point sound source is 85 dB. What is the intensity level 50 m away from the same source
Answer:
425dBExplanation:
Given the intensity level 10 m from a point sound source is 85 dB, then;
L1 = 10m, I1= 85dB ...1
The intensity level 50 m away from the same source cal be calculated using the equivalent expression;
when L2 = 50m, I2 = ? ... 2
Solving equation 1 nad 2;
10m = 85db
50m = x
Cross multiplying;
50 * 85 = 10 * x
10x = 50*85
10x = 4250
Divide both sides by 10
10x/10 = 4250/10
x = 425 dB
Hence, the intensity level 50 m away from the same source is 425dB
A 2-slit arrangement with 60.3 μm separation between the slits is illuminated with 482.0 nm light. Assuming that a viewing screen is located 2.14 m from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side. A. 24.1 mm B. 34.2 mm C. 68.4 mm D. 51.3 mm
Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?
Answer:
The direction of the force will be towards the east
Explanation:
From the question we are told that
The direction of the downward
Generally according to Fleming's right-hand rule(
Thumb - direction of force
Middle finger - direction of current
Index finger - direction of the magnetic field
) and the fact that the earth magnetic field acts from south to north with respect to the four cardinal points then the direction of the force will be toward the east with respect to the four cardinal point on the earth
A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of magnitude 2 T. Assume the magnetic field is uniform. The area of the loop is A = 2 cm2 and the total resistance in the circuit is 7 Ω.
1. Find the maximum induced emf.
e m fmax =
2. Find the maximum current through the bulb.
Imax
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp
1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V
2) The maximum current through the bulb is;
I_max = 1.36 × 10^(-4) A
We are given;
Number of turns; N = 1
Magnitude of magnetic field; B = 2 T
Area; A = 2 cm² = 0.0002 m²
Angular frequency; ω = 143 /min = 2.38 /s
Resistance; R = 7 Ω.
1) Formula for maximum induced EMF is;
EMF_max = NAωB
Plugging in the relevant values gives;
EMF_max = 1 × 0.0002 × 2.38 × 2
EMF_max = 9.52 × 10^(-4) V
2) Formula for maximum current through the bulb is given as;
I_max = EMF_max/R
Plugging in the relevant values;
I_max = (9.52 × 10^(-4))/7
I_max = 1.36 × 10^(-4) A
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Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
A charged particle moving through a magnetic field at right angles to the field with a speed of 25.7 m/s experiences a magnetic force of 2.98 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 4.64 m/s at an angle of 29.2° relative to the magnetic field.
Answer:
The magnetic force would be:
[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]
Explanation:
Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:
F = q v x B
(where the bold represents vectors)
the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:
[tex]F=q\,v\,B\,sin(\theta)[/tex]
Therefore replacing the known quantities for the first case:
[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]
Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:
[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
1.24Mev
Explanation:
Using
E= hc/lambda
= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9
= 1.24Mev
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the other pipe?
Answer:
The length of the longer pipe is L = 2.30 m
Explanation:
Given that:
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.
How long is the other pipe?
From above;
The formula for the frequency of open ended pipes can be expressed as:
[tex]f = \dfrac{nv}{2L}[/tex]
where n = 1 ( since half wavelength exist between those two pipes)
v = 343 m/s and L = 2.08 m
Thus, the shorter pipe produces a frequency of :
[tex]f = \dfrac{1*343}{2*2.08}[/tex]
[tex]f = \dfrac{343}{4.16}[/tex]
[tex]f =82.45 \ Hz[/tex]
Also; we know that the beat frequency was given as 8.0 Hz
Then,
The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz
The lower frequency of the longer pipe = 74.45 Hz
Finally;
From the above equation; make Length L the subject of the formula. Then,
The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]
The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]
The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]
The length of the longer pipe is L = 2.30 m
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
A single-slit diffraction pattern is formed on a distant screen. Assume the angles involved are small. Part A By what factor will the width of the central bright spot on the screen change if the wavelength is doubled
Answer:
If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Explanation:
For a single-slit diffraction, diffraction patterns are found at angles θ for which
w sinθ = mλ
where w is the width
λ is wavelength
m is an integer, m = 1,2,3, ....
From the equation, w sinθ = mλ
For the first case, where nothing was changed
w₁ = mλ₁ / sinθ
Now, If the wavelength is doubled, that is, λ₂ = 2λ₁
The equation becomes
w₂ = mλ₂ / sinθ
Then, w₂ = m(2λ₁) / sinθ
w₂ = 2(mλ₁) / sinθ
Recall that, w₁ = mλ₁ / sinθ
Therefore, w₂ = 2w₁
Hence, If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is
Answer:
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is HELIUM
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
The intensity at a certain distance from a bright light source is 7.20 W/m2 .
A. Find the radiation pressures (in pascals) on a totally absorbing surface and a totally reflecting surface.
B. Find the radiation pressures (in atmospheres) on a totally absorbing surface and a totally reflecting surface.
Answer:
A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa
B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm
Explanation:
A) The formula for radiation pressure for absorbed light is given as;
P_rad = I/c
Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s
Thus;
P_rad = 7.2/(3 × 10^(8))
P_rad.abs = 2.4 × 10^(-8) Pa
Now formula for radiation pressure for reflected light is given as;
P_rad = 2I/c
Thus;
P_rad = (2 × 7.2)/(3 × 10^(8))
P_rad.ref = 4.8 × 10^(-8) Pa
B) Now, 1.013 × 10^(5) Pa = 1 atm
Thus, for the absorbed surface, we have;
P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))
P_rad.abs = 2.369 × 10^(-13) atm
For the reflecting surface, we have;
P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))
P_rad.ref = 4.738 × 10^(-13) atm
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C
Answer:
Objects that are closer together have a stronger force of gravity between them.
Explanation:
For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.
