Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above
Answer:
b. Third Law entropy
Explanation:
Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature" tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.
In the question above, the correct answer is option b.
A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg . Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.
Answer:
602.27 kg
Explanation:
The computation of the largest mass of cargo the balloon can lift is shown below:-
Volume of helium inside the ballon= (4 ÷ 3) × π × r^3
= (4 ÷ 3) × 3.14 × 6.953
= 1406.19 m3
Mass the balloon can carry = volume × (density of air-density of helium)
= 1406.19 × (1.29-0.179)
= 1562.27 kg
Mass of cargo it can carry = Mass it can carry - Mass of structure
= 1562.27 - 960
= 602.27 kg
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Answer:
b. The current in the loop always flows in a counterclockwise direction.
Explanation:
When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.
The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.
This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
Learn more about the magnetic field here:
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A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. Calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one hundredth of a second.
Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
Fₓ t = m ( -v_{xo})
Y axis
t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
sin 30 = v_{yf} / vf
cos 30 = v_{xf} / vf
v_{yf} = vf sin 30
v_{xf} = vf cos 30
sin40 = F_{y} / F
F_{y} = F sin 40
cos 40 = Fₓ / F
Fₓ = F cos 40
let's substitute
F cos 40 t = m ( cos 30 - vₓ₀)
F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
F = 4 sin30 /sin40 v_{f}
F = 3.111 v_{f}
(3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
v_{f} (3,111 cos 40 -4 cos30) = - 80
v_{f} (- 1.0812) = - 80
v_{f} = 73.99
v_{f} = 74 m/s
now we can calculate the force
F = 3.111 73.99
F = 230 N
Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s
Answer:
The time difference is 2.97 sec.
Explanation:
Given that,
Tension = 29 N
Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]
Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]
Length of first section = 2.5 m
Length of second section = 2.8 m
We need to total distance of first pulse
Using formula for distance
[tex]d=2.5+2.5[/tex]
[tex]d_{1}=5.0\ m[/tex]
We need to total distance of second pulse
Using formula for distance
[tex]d=2.8+2.8[/tex]
[tex]d_{2}=5.6\ m[/tex]
We need to calculate the speed of pulse in the first string
Using formula of speed
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]
Put the value into the formula
[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]
[tex]v_{1}=1.24\ m/s[/tex]
We need to calculate the speed of pulse in the second string
Using formula of speed
[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]
[tex]v_{2}=0.80\ m/s[/tex]
We need to calculate the time for first pulse
Using formula of time
[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]
[tex]t_{1}=4.03\ sec[/tex]
We need to calculate the time for second pulse
Using formula of time
[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]
[tex]t_{2}=7\ sec[/tex]
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=t_{2}-t_{1}[/tex]
Put the value into the formula
[tex]\Delta t=7-4.03[/tex]
[tex]\Delta t=2.97\ sec[/tex]
Hence, The time difference is 2.97 sec.
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
The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
Answer:
Speed of the electron is 2.46 x 10^8 m/s
Explanation:
momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]
where [tex]M_{0}[/tex] is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]
where
[tex]\beta = V/c[/tex]
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]
the equation reduces to
1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]
square both sides of the equation, we have
3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]
3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1
2.0625 = 3.0625[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 0.67
β = 0.819
substitute for [tex]\beta = V/c[/tex]
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s
Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet
Answer:
A. magnetic field
Explanation:
The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .
What is a magnetic field ?A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.
As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.
The magnetic field is produced as a result when an electrical current is passed through the conducting wire .
Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .
Learn more about the magnetic fields here, refer to the link given below;
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A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
which category would a person who has an IQ of 84 belong ?
What is the reason for the increase and decrease size of the moon and write down in a paragraph.
Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .
Requried:
What is the magnitude of the charge (in nC) on each bead?
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W
Complete Question
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.
Answer
a. Approximately [tex]5*10^{10}[/tex] fissions per second.
b. Approximately [tex]6*10^{12 }[/tex]fissions per second.
c. Approximately [tex]4*10^{11}[/tex] fissions per second.
d. Approximately [tex]3*10^{12}[/tex] fissions per second.
e. Approximately[tex]3*10^{14}[/tex] fissions per second.
Answer:
The correct option is d
Explanation:
From the question we are told that
The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]
Thus to generated [tex]100 \ J/s[/tex] i.e (100 W ) the rate of fission is
[tex]k = \frac{100}{3.2 *10^{-11} }[/tex]
[tex]k =3*10^{12} fission\ per \ second[/tex]
What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N
Answer:
A. 30.38°
B 5.04N
Explanation:
Using
F= ILBsin theta
2 .55N= 8.4Ax 0.5mx 1.2T x sintheta
Theta = 30.38°
B. If theta is 90°
Then
F= 8.4Ax 0.5mx 1.2x sin 90°
F= 5.04N
Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm from the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?
Answer:
221 lines per millimetre
Explanation:
We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.
Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,
tanθ = 6.09 cm/42.0 cm = 0.145
From trig ratios, cot²θ + 1 = cosec²θ
cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969
sinθ = 1/cosecθ = 1/6.969 = 0.1435
Also, sinθ = mλ/d at the first-order maximum, m = 1. So
sinθ = (1)λ/d = λ/d
Equating both expressions we have
0.1435 = λ/d
d = λ/0.1435
Now, λ = 650 nm = 650 × 10⁻⁹ m
d = 650 × 10⁻⁹ m/0.1435
d = 4529.62 × 10⁻⁹ m per line
d = 4.52962 × 10⁻⁶ m per line
d = 0.00452962 × 10⁻³ m per line
d = 0.00452962 mm per line
Since d = width of grating/number of lines of grating
Then number of lines per millimetre = 1/grating spacing
= 1/0.00452962
= 220.77 lines per millimetre
≅ 221 lines per millimetre since we can only have a whole number of lines.
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Complete question is;
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
13 m
Explanation:
We are given;
Distance between two nearly parallel mirrors; d = 6.5 m
Distance between the face and the nearer mirror; x = 3 m
Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m
Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.
Thus;
Distance of the first reflection of the back of the head in the rear mirror from the object head is;
y' = 2y
y' = 2 × 3.5
y' = 7
The total distance of this image from the front mirror would be calculated as;
z = y' + x
z = 7 + 3
z = 10
Finally, the second reflection of this image will be 10 meters inside in the front mirror.
Thus, the total distance of the image of the back of the head in the front mirror from the person will be:
T.D = x + z
T.D = 3 + 10
T.D = 13m
If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's radius, you would weigh:________
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth.
CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
What is the power P of the eye when viewing an object 61.0 cm away? Assume the lens-to-retina distance is 2.00 cm , and express the answer in diopters.
Answer:
The power of the eye is 51.64 diopters
Explanation:
The power of the eye is given by;
[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]
where;
P is the power of the eye in diopter
f is the focal length of the eye
[tex]d_o[/tex] is the distance between the eye and the object
[tex]d_i[/tex] is the distance between the eye and the image
Given;
[tex]d_o[/tex] = 61.0 cm = 0.61 m
[tex]d_i[/tex] = 2.0 cm = 0.02 m
[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]
Therefore, the power of the eye is 51.64 diopters.
The power P of the eye when viewing an object 61.0 cm away is 51.639D
The power of a lens is a reciprocal of its focal length and it is expressed as:
[tex]P=\frac{1}{f}[/tex]
According to the mirror formula
[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]
where
[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m
[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m
[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]
Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D
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what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor
Answer
The effect is that it Decreases the field current IF and increases slope K1
Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.
Requried:
What is the frequency the horns emit?
Answer: f ≈ 8.5Hz
Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.
For a source moving and a stationary observer, to determine the frequency:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
where:
[tex]f_{0}[/tex] is frequency of observer;
[tex]f_{s}[/tex] is frequency of source;
c is the constant speed of sound c = 340m/s;
[tex]v_{s}[/tex] is velocity of source;
Rearraging for frequency of source:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]
Replacing and calculating:
[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]
[tex]f_{s} = 9.(0.9412)[/tex]
[tex]f_{s} =[/tex] 8.5
Frequency the horns emit is 8.5Hz.
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.
Answer:
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Explanation:
In this exercise you are asked to relate each with the answers
In general, in the optics diagram,
* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point
* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate
* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.
With these statements, let's review the answers
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. Suppose the magnetic field B begins to decrease rapidly in strength
Requried:
What happens to the loop?
1. The loop is pushed to the left, toward the magnetic field.
2. The loop doesn’t move.
3. The loop is pushed downward, towards the bottom of the page.
4. The loop will rotate.
5. The loop is pushed upward, towards the top of the page.
6. The loop is pushed to the right, away from the magnetic field
Answer:
. The loop is pushed to the right, away from the magnetic field
Explanation
This decrease in magnetic strength causes an opposing force that pushes the loop away from the field
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?
Explanation:
Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,
[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]
[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]
[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]
[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]
after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,
[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]
[tex]\qquad \Phi_{B, f}=0[/tex]
(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,
[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]
[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]
[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]
(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb
(b) The induced EMF in the coil is 0.583 mV
Flux and induced EMF:Given that the coil has N = 250 turns
and an area of A = 14cm² = 1.4×10⁻³m².
It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T
(a) The flux passing through the coil is given by:
Ф = NBAcosθ
where θ is the angle between area vector and the magnetic field
The area vector is perpendicular to the plane of the coil.
So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil
So the initial flux is:
Φ = NABcos0° = NAB
Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb
Ф = 1.75 × 10⁻⁵ Wb
Finally, θ = 90°, and since cos90°, the final flux through the coil is 0
(b) The EMF induced is given by:
E = -ΔФ/Δt
E = -(0 - 1.75 × 10⁻⁵)/0.030
E = 0.583 × 10⁻³ V
E = 0.583 mV
Learn more about magnetic flux:
https://brainly.com/question/15359941?referrer=searchResults
Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Answer:
Explanation:
magnetic field due to an infinite current carrying conductor
B = k x 2I / r where k = 10⁻⁷ , I is current in conductor and r is distance from wire
putting the given data
B = 10⁻⁷ x 2 x 100 / 8
= 25 x 10⁻⁷ T .
B )
earth's magnetic field = .5 gauss
= .5 x 10⁻⁴ T
= 5 x 10⁻⁵ T
percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100
= 5 %
C )
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the device
Answer: R=24.2Ω
Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
[tex]P=\frac{V^{2}}{R}[/tex]
The resistance of the system is:
[tex]P=\frac{V^{2}}{R}[/tex]
[tex]R=\frac{V^{2}}{P}[/tex]
[tex]R=\frac{110^{2}}{500}[/tex]
R = 24.2Ω
For the device, resistance is 24.2Ω.
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.
Answer:
8 seconds
Explanation:
Answer:
Explanation:
Going up
Time taken to reach maximum height= usin∅/g
=3 secs
Maximum height= H+[(usin∅)²/2g]
=80+[(60sin30)²/20]
=125 meters
Coming Down
Maximum height= ½gt²
125= ½(10)(t²)
t=5 secs
