Answer:
Option A & B
Explanation:
This is cause the compass needle is magnetized anointed in such a way that it responds to magnetic field strength.
Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 1.00×104 hours, on the average, and costs $ 12.0 , whereas the incandescent bulb costs only 76.0 ¢, but lasts just 750 hours. The study assumed that electricity cost 9.00 ¢ per kWh and that the bulbs were on for 4.0 h per day.
Answer:
The resistance is 626.0 Ω.
Explanation:
Given that,
Power of compact bulb= 100 W
Power of incandescent bulb = 23.0 W
Time [tex]t= 1.00\times10^{4}\ hours[/tex]
What is the resistance of a “100 W”fluorescent bulb? (Remember the actual rating is only 23W of powerfor a 120V circuit)
We need to calculate the resistance of the bulb
Using formula of power
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]R=\dfrac{V^2}{P}[/tex]
Where, P = power
R = resistance
V = voltage
Put the value into the formula
[tex]R=\dfrac{120^2}{23}[/tex]
[tex]R=626.0\ \Omega[/tex]
Hence, The resistance is 626.0 Ω.
Which motion maps show an object in uniform circular motion? Check all that apply. V W X Y
The correct options are V, W, and Y
Uniform Circular Motion is the motion when an object moves in a circular path with constant speed (uniform speed). In such a motion,
1- The position keeps on changing,
2- The speed remains constant
Now checking all the option
V- It shows the uniform circular motion since all the direction is uniform,
W-It shows the uniform circular motion since all the direction is uniform,
X-Direction are different,
Y-It shows the uniform circular motion since all the direction is uniform,
Hence, all the above options,
The correct options are V, W, and Y
For more about the uniform circular motion follow the link below,
https://brainly.com/question/12920060
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
Answer : The energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant = 10973731.6 m⁻¹
[tex]n_f[/tex] = Higher energy level = 3
[tex]n_i[/tex]= Lower energy level = 2
Putting the values, in above equation, we get:
[tex]\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\lambda=6.56\times 10^{-7}m[/tex]
Now we have to calculate the energy.
[tex]E=\frac{hc}{\lambda}[/tex]
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = [tex]6.56\times 10^{-7}m[/tex]
Putting the values, in this formula, we get:
[tex]E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}[/tex]
[tex]E=3.03\times 10^{-19}J[/tex]
Therefore, the energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]
A rifle with a weight of 25 N fires a 4.0 g bullet with a speed of 290 m/s.
(a) Find the recoil speed of the rifle.
Your response is off by a multiple of ten. m/s
(b) If a 750 N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.
m/s
Answer:
Explanation:
Given
Weight of rifle [tex]W=25\ N[/tex]
mass of rifle [tex]M=\frac{25}{10}=2.5\ kg[/tex]
mass of bullet [tex]m=4\ gm[/tex]
speed of bullet [tex]u=290\ m/s[/tex]
As there is no net force on the bullet-rifle system therefore momentum is conserved
[tex]0=Mv+mu[/tex]
[tex]v=-\frac{mu}{M}[/tex]
[tex]v=-\frac{4\times 10^{-3}\times 290}{2.5}[/tex]
[tex]v=-0.464\ m/s[/tex]
i.e. opposite to the direction of bullet speed
(b)Weight of Man [tex]=750\ N[/tex]
Combined weight of man and rifle[tex]=750+25=775\ N[/tex]
mass of man-rifle system [tex]M'=\frac{775}{10}=77.5\ kg[/tex]
Now man and rifle combinedly recoil
therefore
[tex]0=(M')v '+mu [/tex]
[tex]v'=-\frac{mu}{M'}[/tex]
[tex]v'=-\frac{4\times 10^{-3}\times 290}{77.5}[/tex]
[tex]v'=0.0149\ m/s[/tex]
A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an angle of 25° with the horizontal.
a. What are the horizontal and vertical components of the applied force?
b. What is the magnitude of each of the forces? i) Applied Force ii) Weight ii) Normal Force iii) Frictional Force
c. How much work is done by each of the forces? i) WApplied Force ii) WWeight iii) WNormal Force iv) WFrictional Force
d. What is the total amount of work done on the object?
e. What is the coefficient of friction of the crate on the floor?
Answer:
A) Fx = 45.3N Fy = 21.1N
B) Fm = 45.3N R = 196N Ff = 35.4N
C) i) 0KJ ii) 11.5KJ iii) 11.5KJ iv) 0KJ
D) 23KJ
E) 0.047
Explanation:
a) Horizontal component of the force Fx = 50cos 25° = 45.3N
Vertical component = 50sin25° = 21.1N
b) Magnitude of the applied force = moving force Fm = Fx = 50cos25° =45.3N
Magnitude of the weight = mg = 20×9.8 = 196N
Magnitude of normal force which is the reaction is equal to the weight = mg = 20×9.8 = 196N
Frictional force = moving force = 50cos45° = 35.4N
c) since workdone = Force done × perpendicular distance in the direction of the force
- workdone on the moving force is 0Joules since it has no perpendicular distance
- workdone on weight is the weight × distance = (20×9.8)×12 = 11.5KJ
= work done on normal force = workdone by the weight = 11.5KJ
Workdone on the frictional force is is 0Joules since the force is along the horizontal (no perpendicular distance)
d) the total work done = work done by Applied Force +Weight + Normal Force + Frictional Force= 0+11.5+11.5 = 23KJ
e) the coefficient of friction = moving force / normal reaction = 45.3/196 = 0.047
Explanation:
A.
Horizontal component, Fy = F * cos(a)
= 50cos25
= 50 * 0.91
= 45.32 N
Vertical component,Fx = F * sin(a)
= 50sin25
= 50 * 0.42
= 21.13 N
B.
Applied force = 50 N
Weight,W = m * g
= 20 * 9.81
= 196.2 N
Normal force, N = W - Fx
= 200 - 21.13
= 179 N
Frictional force = Fy
= 45.32 N.
C.
Workdone by Normal and Weight forces are = 0, because they both act perpendicular to the movement.
Workdone by friction = Workdone by applied forces
= force * distance
= 12 * 45.32
= 543.84 J
D.
Total amount of work done on the crate = 0 (the movement with a constant speed).
E.
The coefficient of friction, u = Fy/(W - Fx)
= 45.5/(196.2 - 21.13)
= 0.26
A 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average force required to stop the player in 0.65 second?
(1) 120 N
(2) 290 N
(3) 690 N
(4) 920 N
Answer:
(3) 690 N
Explanation:
Force: This is the product of mass and acceleration of a body. The S.I unit of force is Newton(N).
The formula for force is given as,
F = ma..................... Equation 1
Where F = force, m = mass, a = acceleration.
Also,
a = (v-u)/t................... Equation 2
Where v = Final velocity, u = initial velocity, t = time.
Given: u = 6.0 m/s, v = 0 m/s (bring to stop), t = 0.65 s.
Substitute into equation 2
a = (0-6)/0.65
a = -6/0.65
a = -9.23 m/s²
Also given: m = 75 kg
Substitute again into equation 1
F = 75(-9.23)
F = -692.25 N
The negative sign tells that the force oppose the motion of the player
F ≈ 690 N
Hence the right option is (3) 690 N
The magnitude of the average force required to stop the player in 0.65 second is 690 N. And option (3) is correct.
Given data:
The mass of hockey player is, m = 75 kg.
The speed of hockey player is, v = 6.0 m/s.
The reaction time is, t = 0.65.
Use the concept of impulse - momentum theorem to obtain the value of average force acting on the hockey player as,
As per the impulse - momentum theorem,
[tex]F = \dfrac{mv}{t}[/tex]
Here, F is the average force exerted on the hockey player.
Solving as,
[tex]F = \dfrac{75 \times 6}{0.65 }\\\\F \approx 690 \;\rm N[/tex]
Thus, the magnitude of the average force required to stop the player in 0.65 second is 690 N. And option (3) is correct.
Learn more about the impulse - momentum theorem here:
https://brainly.com/question/24744548
A(n) _____ is a procedure for using ultrasonic sound waves to create a picture of an embryo or fetus.
Answer:
Sonogram
Explanation:
A sonogram is a medical diagnostic method that is commonly uses sound waves in order to generate images of any tissues, organs, or some other internal structures inside the human body. This sonogram does not use any kind of harmful radiations like the X-rays. These are sound waves of high frequencies.
This is widely used by doctors for health checkups. The other name for sonogram is ultrasound.
A cannonball is shot out of a long cannon and then a short cannon with the same amount of force. The short cannon cannonball has Group of answer choices
more mometum, velocity, and impulse.
more momentum only.
more impulse only.
less mometum.
more velocity only.
Answer:
The correct answer is less momentum.
Explanation:
The momentum of an object is directly linked to the force exerted on that object. In a longer barrel, the shot cannonball will have a greater impulse changing to occur a larger momentum and therefore a higher speed will be applied to the bullet, contrary to what would happen in a shorter barrel.
Final answer:
The cannonball shot from a short cannon does not have more momentum, velocity, or impulse by virtue of the cannon's length alone, since momentum and impulse depend on both the force applied and the time the force acts on the cannonball.
Explanation:
If a cannonball is shot from both a long and a short cannon with the same amount of force, we need to understand that force, momentum, and impulse are interrelated but distinct concepts in physics. The product of an object's mass and velocity is its momentum.
Impulse is the change in momentum, often calculated as force multiplied by the time the force is applied. Because the question states that the force is the same, the impulse given to the cannonball would be identical regardless of the cannon length, assuming the force is applied over the same time period.
However, a longer cannon might allow the cannonball more time to accelerate, potentially resulting in greater final velocity and therefore more momentum, since momentum is mass times velocity. Conversely, a shorter cannon may allow less time for the force to act, possibly resulting in a lower velocity and, by extension, less momentum.
Thus, the cannonball from the short cannon does not necessarily have more velocity only, as it depends on the time the force acts on the cannonball.
In summary, the answer choice would be that the short cannon cannonball does not categorically have more momentum, velocity, or impulse based solely on the length of the cannon.
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
There is an omission of some sentences in the question which affects the answering of question B and C, so we will based the omission of the sentences on assumption in order to solve the question that falls under it.
NOTE: The omitted sentences are written in bold format
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
Answer:
a) [tex]V_{avg} = 20.5m/s[/tex]
b) 9.76 × 10⁻⁴s
c) 247.9 N
d) 5.8 N
Explanation:
Given that;
Initial speed [tex](V_i)[/tex] = 0
Final speed [tex](V_f)[/tex] = 41 m/s
Distance (d) = 0.002
mass (m) = 0.059 kg
g = 9.8 m/s²
a)
The average speed of the ball can be calculated as;
[tex]V_{avg} = \frac{V_i+V_f}{2}[/tex]
[tex]V_{avg} = \frac{0+41}{2}[/tex]
[tex]V_{avg} = 20.5m/s[/tex]
b)
The time elapsed can be calculated by using the second equation of motion which is given as:
[tex]S=(\frac{V_i+V_f}{2})t[/tex]
If we make time (t) the subject of the formula; we have:
[tex](V_i+V_f)t=2S[/tex]
[tex]t= (\frac{2S}{V_I+V_f})[/tex]
[tex]=\frac{2(0.02)}{41+0}[/tex]
[tex]=\frac{0.04}{41}[/tex]
= 0.000976
=9.76 × 10⁻⁴s
c)
the magnitude of the average force (F) exerted by the wall on the bal dring contact can be determined using;
Force (F) = mass × acceleration
where acceleration [tex](a)= \frac{Vo}{t}[/tex]
[tex]\frac{41}{0.00976}[/tex]
acceleration (a) = 4200.82 m/s²
F = m × a
= 0.059 × 4200.82
= 247.85
≅ 247.9 N
d)
the magnitude of the gravitational force of the Earth on the ball
Force (F) = mass (m) × gravity (g)
= 0.059kg × 9.8 m/s²
= 5.782 N
≅ 5.8 N
The average speed of the tennis ball during contact with the wall is zero, and without the time of contact, we cannot determine the time elapsed or the average force exerted by the wall. However, the gravitational force on the ball is 0.5782 N.
Explanation:The question relates to the change in momentum and the forces involved when a tennis ball bounces off a wall. Specifically, a tennis ball with a mass of 0.059 kg is hit at a speed of 41 m/s, bounces off a wall, and comes back at the same speed. To tackle the posed questions, it is essential to apply concepts from Newton's laws of motion and the conservation of momentum.
Part AThe average speed of the ball during contact is zero since the speed decreases uniformly from 41 m/s to zero.
Part BWithout the time of contact with the wall, this cannot be determined. Previous examples of collisions show time of contact can vary, so it must be provided to answer this part of the question.
Part CTo calculate the magnitude of the average force exerted by the wall on the ball, we would need the time of contact with the wall. Since it is not given, this cannot be calculated accurately.
Part DThe magnitude of the gravitational force of the Earth on the ball is calculated as the product of the mass of the ball and the acceleration due to gravity (9.8 m/s²), which is 0.059 kg * 9.8 m/s² = 0.5782 N.
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.6 vibrations in 27.9 s. Also, a given maximum travels 450 cm along the rope in 11.3 s. What is the wavelength? Answer in units of cm.
Answer:
[tex]\lambda = 25.79\ cm[/tex]
Explanation:
given,
Wave vibrates = 37.6
time = 27.9 s
maximum distance travel = 450 cm
time = 11.3 s
wavelength = ?
frequency of wave
[tex]f=\dfrac{37.6}{27.9}[/tex]
f = 1.35 Hz
Speed of wave
[tex]v = \dfrac{450}{11.3}[/tex]
v = 39.82 cm/s
wavelength of wave
v = fλ
[tex]\lambda =\dfrac{v}{f}[/tex]
[tex]\lambda =\dfrac{34.82}{1.35}[/tex]
[tex]\lambda = 25.79\ cm[/tex]
Hence, wavelength of the wave is equal to 25.79 cm.
Final answer:
The wavelength of a wave can be calculated using the speed of the wave and its frequency. In this case, the wavelength is 29.50 cm.
Explanation:
Wavelength: The wavelength of a wave is the distance between two similar points on the medium that have the same height and slope. In this case, the wavelength can be found using the formula: λ = v/f, where v is the speed of the wave and f is the frequency.
Given Data: Number of vibrations = 37.6, Time = 27.9 s, Distance traveled = 450 cm, Time = 11.3 s.
Calculation: Speed of the wave = Distance/Time = 450 cm / 11.3 s = 39.82 cm/s. Frequency = Number of vibrations / Time = 37.6 / 27.9 = 1.35 Hz. Therefore, Wavelength = 39.82 cm / 1.35 Hz = 29.50 cm.
Blood is accelerated from rest to 25.00 cm/s in a distance of 2.10 cm by the left ventricle of the heart. How long does the acceleration take? (To solve this problem, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.)
Answer:
Time taken, t = 0.16 seconds
Explanation:
Given that,
Initial speed of blood, u = 0
Final speed of blood, v = 25 cm/s = 0.25 m/s
Distance, d = 2.1 cm = 0.021 m
Using first equation of motion as :
v = u + at
u = 0
[tex]v=at\\0.25 =at[/tex]
Let t is the time taken. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
[tex]d=\dfrac{1}{2}at\times t[/tex]
[tex]t=\dfrac{2d}{at}[/tex]
Since, at = 0.25
So,
[tex]t=\dfrac{2\times 0.021}{0.25}[/tex]
t = 0.16 seconds
So, the time taken by the blood to accelerate is 0.16 seconds.
A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires and the roadway is 0.400. A race car with mass 1200 kg rounds the curve with the maximum speed to avoid skidding. consider friction when solving for a, b, and c.
a) As the car rounds the curve, what is the normal force exerted on it by the road?
b) What is the car's radial acceleration?
c) What is the car's speed?
d) In the case of a banked curve with friction, which of the following forces contribute to the centripetal (inward) acceleration: the frictional force, the normal force, and/or the gravity? and why?
Answer:
See attachment below
Explanation:
The normal force exerted on the car by the curved road is 13,139.7 N.
The radial acceleration of the car is 6.56 m/s².
The speed of the car is 21.43 m/s.
In the case of a banked curve with friction, the centripetal acceleration is increased by normal force and frictional force since it prevents the car from skidding.
The given parameters;
radius of the curved path, r = 70 mbanking angle, θ = 12⁰coefficient of friction, μ = 0.4 mass of the race car, m = 1200 kgThe normal force exerted on the car by the curved road is calculated as follows;
[tex]\Sigma F_y = 0\\\\Ncos(\theta) - \mu_s Nsin(\theta) - mg = 0\\\\Ncos(\theta) - \mu_s Nsin(\theta) = mg\\\\N(cos\theta \ - \ \mu_s sin(\theta)) = mg\\\\N = \frac{mg }{cos\theta - \mu_s sin(\theta)} \\\\N = \frac{(1200 \times 9.8)}{cos (12) - \ 0.4\times sin(12)} \\\\N = 13,139.7 \ N[/tex]
The radial acceleration of the car is calculated as follows;
[tex]\Sigma F_x = ma_c\\\\ma_c = Nsin(\theta) + \mu_s N cos(\theta)\\\\a_c = \frac{Nsin(\theta) + \mu_s N cos(\theta)}{m} \\\\a_c = \frac{(13, 139.7)sin(12) \ + \ 0.4 \times 13,139.7cos(12)}{1200} \\\\a_c = 6.56 \ m/s^2[/tex]
The car's speed is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\v^2 = a_c r\\\\v = \sqrt{a_c r} \\\\v = \sqrt{6.56 \times 70} \\\\v = 21.43 \ m/s[/tex]
In the case of a banked curve with friction, the centripetal acceleration is increased by the following;
[tex]ma_c = Nsin(\theta) + F_s\\\\ma_c = Nsin(\theta) + \mu_s Ncos (\theta)[/tex]
where;
[tex]F_s[/tex] is the frictional forceN is the normal forceThus, In the case of a banked curve with friction, the centripetal acceleration is increased by normal force and frictional force since it prevents the car from skidding.
Learn more here:https://brainly.com/question/14209158
A piano tuner hears a beat every 1.80 s when listening to a 272.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Answer:
272.56 or 271.44
Explanation:
[tex]Fbeat=\frac{1}{Tbeat} \\=\frac{1}{1.8} \\=0.56 Hz[/tex]
[tex]Fbeat=abs(f1-f2)\\0.56=abs(272-f2)\\\\f2=272.56\\f2=271.44[/tex]
In the video, the torque due to the mass of the plank is used in the calculations. For this question, ignore the mass of the board. Rank, from largest to smallest, the mass m needed to keep the board from tipping over. To rank items as equivalent, overlap them.
Answer:
D, A & B, C, E
Explanation:
In each problem, sum the torques about the edge of the table.
A) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (1.5 m) = 0
m = 100 kg
B) ∑τ = Iα
mg (-0.75 m) + (100 kg) g (0.75 m) = 0
m = 100 kg
C) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (0.75 m) = 0
m = 50 kg
D) ∑τ = Iα
mg (-1.5 m) + (200 kg) g (1.5 m) = 0
m = 200 kg
E) ∑τ = Iα
mg (-1.5 m) + (100 kg) g (0.5 m) = 0
m = 33.3 kg
A football player runs the pattern given in the drawing by the three displacement vectors , , and . The magnitudes of these vectors are A = 4.00 m, B = 13.0 m, and C = 19.0 m. Using the component method, find the (a) magnitude and (b)direction of the resultant vector + + . Take to be a positive angle.
Answer:
Explanation:
check the attached for the solution
If an aircraft is loaded 90 pounds over maximum certificated gross weight and fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
15 gallons.
Gallons = (Pounds) ÷ (Pounds ÷ Gallons)
Explanation:
To bring an aircraft's weight under the maximum certificated gross weight by 90 pounds, 15 gallons of gasoline should be drained.
If an aircraft is loaded with 90 pounds over the maximum certificated gross weight and needs to be brought within the weight limit by draining fuel, we need to calculate how much fuel, in pounds, should be drained. Gasoline's weight can vary slightly with temperature, but for general aviation purposes, it's usually taken to be about 6 pounds per gallon.
Hence, to remove 90 pounds of weight, the amount of fuel that should be drained would be 90 divided by 6.
Performing the calculation, 15 gallons of fuel should be drained to remove 90 pounds of weight from the aircraft and bring it back to or below the maximum certificated gross weight.
What are constellations?
1. Apparent groupings of stars and planets visible on a given evening
2. Groups of galaxies gravitationally bound and close together in the sky
3. Groups of stars gravitationally bound and appearing close together in the sky
4. Groups of stars making an apparent pattern in the celestial sphere
5. Ancient story boards, useless to modern astronomers
Answer:
Option 4
Explanation:
A constellation can be defined as that region formed by the stars in such a way that the formation by the group of stars in that area appear to seem an imaginary pattern of some mythological creature, animal, god or some inanimate object formed apparently.
Thus in accordance with the above definition a constellation is a group of stars that forms some apparent pattern in the celestial sphere.
In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 2.30 104-m length of wire was let out by the space shuttle Atlantis to generate a motional emf. The shuttle had an orbital speed of 7.50 103 m/s, and the magnitude of the earth's magnetic field at the location of the wire was 5.40 10-5 T. If the wire had moved perpendicular to the earth's magnetic field, what would have been the motional emf generated between the ends of the wire?
emf = 9.3 x 10³
Explanation:
When a conductor moves in the magnetic field , the emf is generated across its ends . Which can be calculated by the relation
emf ξ = B x l x v
here B is the magnetic field strength , l is the length of conductor and v is its velocity .
In our question B = 5.4 x 10⁻⁵ T
l = 2.30 x 10⁴ m and v = 7.5 x 10³
Thus ξ = 5.4 x 10⁻⁵ x 2.30 x 10⁴ x 7.5 x 10³ = 9.3 x 10³ Volt
Fiora starts riding her bike at 18 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 69 miles takes her 4.5 hours. For how long did she travel at 18 mi/h?
Answer:
t = 2.5 hours
Explanation:
given,
speed of the bike for t time= 18 mi/h
final speed of the bike after t time = 12 mi/h
total distance, D = 69 miles
total time, T= 4.5 hour
time for which speed of the bike is 18 mi/h = ?
we know distance = speed x time
now,
18 x t + 12 (4.5 - t) = 69
6 t + 54 = 69
6 t = 15
t = 2.5 hours
The bike was at the speed of 18 mi/h for 2.5 hours.
Under what circumstances can energy level transitions occur?
Energy level transitions in physics occur when an atom or molecule absorbs or emits electromagnetic radiation. The circumstances of these transitions depend on the system being studied.
Explanation:Energy level transitions in physics typically occur when an atom or a molecule absorbs or emits electromagnetic radiation. When an electron within an atom moves from a lower energy level to a higher energy level, it absorbs energy. Conversely, when an electron moves from a higher energy level to a lower energy level, it emits energy in the form of electromagnetic radiation.
These transitions can occur under various circumstances, such as when an atom is excited by heat or light, or when an electron interacts with another particle. For example, in the hydrogen atom, transitions between different energy levels give rise to the absorption and emission of specific wavelengths of light, leading to the existence of discrete spectral lines. It is important to note that the exact circumstances of energy level transitions depend on the specific system being studied, such as atoms, molecules, or subatomic particles.
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Energy level transitions in atoms occur when an electron absorbs or emits energy, typically through photon interactions, due to electronic rearrangements.
Energy level transitions in atoms happen when electrons change their energy states. These transitions can occur under several circumstances:
Absorption of Photons: Electrons can move to higher energy levels by absorbing photons with specific energy levels corresponding to the energy difference between the levels. This is observed in absorption spectra.
Emission of Photons: Electrons release energy in the form of photons when transitioning from higher to lower energy levels. These emitted photons produce emission spectra.
Electron Collisions: In certain situations, such as in a plasma, collisions between electrons and other particles can excite electrons to higher energy states, leading to subsequent de-excitation and photon emission.
External Energy Sources: External sources like electrical discharges or high-temperature environments can energize electrons, causing transitions between energy levels.
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A 5.67 gram coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of the static friction is 0.100, how far from the center of the record can the coin be placed without having it slip off?
Answer:
81 mm from the center
Explanation:
5.67g = 0.00567 kg
33.3 revolutions per minute = 33.3 (rev/min) * 2π (rad/rev) * (1/60) (min/sec) = 3.487 rad/s
The weight of the coin is product of mass and gravitational acceleration g = 9.81m/s2
W = mg = 0.00567 * 9.81 = 0.0556 N
Which is also the normal force of the record acting back on the coin to balance it.
Them the static friction is product of normal force and friction coefficient
[tex]F_f = N\mu = 0.0556*0.1 = 0.00556 N[/tex]
For the coin to NOT slip off, its centripetal force should at most be equal to the static friction
[tex]F_c = 0.00556[/tex]
[tex]a_cm = 0.00556[/tex] Newton's 2nd law
[tex]a_c = 0.00556 / 0.00567 = 0.981 m/s^2[/tex]
The centripetal acceleration is the product of squared angular velocity and radius of circular motion
[tex]a_c = \omega^2r[/tex]
[tex]r = \frac{a_c}{\omega^2} = \frac{0.981}{3.487^2} = 0.081m[/tex] or 81 mm
Indicate whether each of the following statements is true or false. 1. The molecules in hot air move at the same speed as in cold air, but there are more molecules so they feel hotter. 2. In a coal-fired power plant, the types of energy from start-to-finish are: electrical, mechanical, thermal, and chemical. 3. In a coal-fired power plant, the approximate percentage of original energy in the coal lost to heat is 10%.
Answer:
1) False
2) False (chemical, thermal, mechanical and electrical)
3) False
Explanation:
1.
We have the expression for the root mean square velocity of the molecules of gases as:
[tex]v_{rms}=\sqrt{\frac{3.k_b.T}{M} }[/tex]
where:
[tex]k_b=[/tex] Boltzmann constant
[tex]T=[/tex] temperature of the gas
[tex]M=[/tex] molecular mass of the gas
2.
In a coal-fired powered the types of energy form start to finish can be given as;
At first the chemical energy of the coal gets converted into heat energy after the process of combustion.Then next, this heat is utilized for generating high pressure steam which drives the turbine converting the heat energy into the mechanical energy.Then this rotational motion of turbine shaft is coupled with the armature of the alternator to convert the mechanical energy into electrical energy.3.
In a coal fired power plant the approximate percentage of original energy in the coal lost to heat is approximately 60%.
Two charges that are separated by one meter exert 1-N forces on each other. If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be______
Final answer:
If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be 16 N.
Explanation:
In electrostatics, the force between two charged particles is given by Coulomb's law. The formula to calculate the force is F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.
Using this formula, if the distance between the charges is decreased from 1 meter to 25 centimeters, the force on each charge will increase. The force is inversely proportional to the square of the distance, so when the distance is reduced to 0.25 meters (25 centimeters), the force will be 16 times stronger. Therefore, the force on each charge will be 16 N.
What is the angular diameter (in arcseconds) of an object that has a linear diameter of 75 cm and a distance of 2 km?
Answer:
The angular diameter is 77.35 arc-seconds.
Explanation:
The angular diameter, as shown in the figure, is the angle [tex]x[/tex] subtended by the the diameter of the object.
Before we do the calculation, we first convert everything to meters.
The diameter of the of the object in meters is
75cm =0.75m,
and the distance to the object in meters is
2 km = 2000 m.
Now, from trigonometry we get:
[tex]tan (\frac{x}{2} )= \dfrac{radius}{length} = \dfrac{0.75/2}{2000} \\\\\dfrac{x}{2} = tan^{-1}(\dfrac{0.75/2}{2000})\\\\\dfrac{x}{2}= 0.0107\\\\\boxed{x= 0.0215^o}[/tex]
and since 1 degree = 3600 arc-seconds, [tex]x[/tex] in arc-seconds is
[tex]x= 0.0215*3600 \\\\\boxed{x= 77.35''}[/tex]
A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular speed. Answer in units of rad/s.
Answer:
[tex]\omega=0.12\frac{rad}{s}[/tex]
Explanation:
In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:
[tex]\omega=\frac{2\pi}{T}(1)[/tex]
Here T is the period, that is, the time taken to complete onee revolution:
[tex]T=\frac{2\pi r}{v}(2)[/tex]
Replacing (2) in (1):
[tex]\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}[/tex]
The angular speed of the racing car is approximately 0.1208861 rad/s.
The angular speed of the racing car can be calculated using the formula that relates linear speed (v) to angular speed and the radius (r) of the circular path, which is:
[tex]\[ v = r \cdot \omega \][/tex]
Given that the linear speed (v) of the car is 19.1 m/s and the radius (r) of the circular track is 158 m, we can solve for the angular speed (Ï) as follows:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{19.1 \text{ m/s}}{158 \text{ m}} \][/tex]
[tex]\[ \omega =\frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = 0.1208861 \text{ rad/s} \][/tex]
Therefore, the angular speed of the racing car is approximately 0.1208861 rad/s.
To express this in a more simplified fraction, we can write:
[tex]\[ \omega \ = \frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \times \frac{10}{10} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{1910}{15800} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
Since 191 and 1580 do not have any common factors other than 1, this fraction is already in its simplest form. Thus, the final answer for the angular speed of the racing car is:
[tex]\[ \ {0.1208861 \text{ rad/s}} \][/tex]
An electron beam in an oscilloscope is deflected by the electric field produced by oppositely charged metal plates. If the electric field between the plates is 2.07 x 105 N/C directed downward, what is the force on each electron when it passes between the plates?
Answer:
Explanation:
Given
Electric Field Strength [tex]E_0=2.07\times 10^5\ N/C[/tex]
Charge on electron [tex]q=1.6\times 10^{-19}\ C[/tex]
Force on any charged particle in an Electric field is given by
[tex]Force=charge\ on\ electron\times Electric\ Field\ strength [/tex]
[tex]F=1.6\times 10^{-19}\times 2.07\times 10^5[/tex]
[tex]F=3.312\times 10^{-14}\ N[/tex]
as the electric field is pointing downward therefore force will be acting downward on a positively charged particle but opposite for an electron i.e. in upward direction
A pair of oppositely charged parallel plates is separated by 5.38 mm. A potential difference of 623 V exists between the plates. What is the strength of the electric field between the plates? The fundamental charge is 1.602 × 10−19 . Answer in units of V/m.
Answer:
115799.256V/m
Explanation:
Given V = 623v, d = 5.38mm = 0.00538m
Strength of electric field E = V/d
E = 623/0.00538
= 115799.256V/m
The wavelength of green light from a traffic signal is centered at 5.20*10^-5cm. Calculate the frequency.
Answer:
5.8×10^11Hz
Explanation:
Frequency is the ratio of the speed of the light wave to its wavelength.
Frequency (f)= velocity(v)/Wavelength (¶)
Given wavelength = 5.2×10^-5cm = 5.2×10^-3m (converted to meters)
Velocity of light = 3×10^8
Substituting the values given in the formula we have
Frequency = 3×10^8/ 5.2×10^-3
Frequency = 5.8×10^11Hz
The frequency of the green light from the traffic signal is 5.77 × 10¹⁴ Hertz.
Given the data in the question;
Wavelength; [tex]\lambda = 5.20*10^{-5}cm = 5.20*10^{-7}m[/tex]
Frequency; [tex]f = \ ?[/tex]
To determine the frequency of the green light, we use the expression for the relations between wavelength, frequency and speed of light.
[tex]\lambda = \frac{c}{f} \\[/tex]
Where [tex]\lambda[/tex] is wavelength, f is frequency and c is speed constant ([tex]3* 10^8m/s[/tex])
We substitute the values into the equation
[tex]5.20*10^{-7}m = \frac{3*10^8m/s^}{f} \\\\f = \frac{3*10^8m/s^}{5.20*10^{-7}m}\\\\f = 5.77 * 10^{14}s^{-1}\\\\f = 5.77 * 10^{14}Hz[/tex]
Therefore, the frequency of the green light from the traffic signal is 5.77 × 10¹⁴ Hertz.
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An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be A) Th/2 B) 7V2. C) T/4. D) 4T E) TI2
Answer: Option (B) is the correct answer.
Explanation:
Expression for centripetal acceleration is as follows.
a = [tex]r \omega^{2}[/tex] .......... (1)
Also, we know that
[tex]\omega = \frac{2 \pi}{T}[/tex] ........... (2)
Putting the value from equation (2) into equation (1) as follows.
a = [tex]r (\frac{2 \pi}{T})^{2}[/tex]
= [tex]r \frac{2 (\pi)^{2}}{4}[/tex]
As, [tex]a \propto \frac{1}{T^{2}}[/tex]
a = [tex]\frac{k}{T^{2}}[/tex]
or, [tex]aT^{2} = k[/tex]
Now, we will reduce a to [tex]\frac{a}{2}[/tex]. So, new value of [tex]T^{2}[/tex] will be equal to [tex]2T^{2}[/tex].
Therefore, value of new period will be as follows.
[tex]T' = \sqrt{2T^{2}}[/tex]
= [tex]\sqrt{2}T[/tex]
Thus, we can conclude that the new period is equal to [tex]T \sqrt{2}[/tex].
The new period should be [tex]T\sqrt{2}[/tex]
Calculation of new period:The expression for centripetal acceleration should be [tex]a = r\omega^2[/tex]
Now
we know that [tex]w = 2\pi \div T[/tex]
Now here we put the values
[tex]a = r(2\pi\div T)^2\\\\= r (2(\pi)^2\div 4\\\\Since\ a \alpha \frac{1}{T^2}\\\\ aT^2 = K[/tex]
Now here we have to decrease to a by 2. So the new T value should be [tex]2T^2[/tex]
So, the new period should be
[tex]T = \sqrt{2T^2}\\\\ = \sqrt{2T}[/tex]
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A(n) _____ satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Answer:
The correct answer is a Low earth orbit.
Explanation:
A low earth orbit can be understood as an earth orbit with an altitude of 1,000 miles or less. It is a satellite sustem that employs many satelites, in fact, most man-made objects that are currently in outer-space are part of this low earth orbit. (LEO).
The most famous LEO satellite system is the one from planet earth. Almost every space flight that human beings have ever done are done in LEO, and every spacial station is located in this zone.
In conclusion, A low earth orbit satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Answer:A(n) low Earth Orbit (LEO) satellite system employs many satellites, each in an orbit at an altitude of less than 1,000 miles.
Explanation:A low Earth orbit (LEO) is an Earth-centered orbit with an altitude of 2,000 km (1,200 mi) or less (approximately one-third of the radius of Earth), or with at least 11.25 periods per day (an orbital period of 128 minutes or less) and an eccentricity less than 0.25. Most of the manmade objects in outer space are in LEO.
There is a large variety of other sources that define LEO in terms of altitude. The altitude of an object in an elliptic orbit can vary significantly along the orbit. Even for circular orbits, the altitude above ground can vary by as much as 30 km (19 mi) (especially for polar orbits) due to the oblateness of Earth's spheroid figure and local topography