a) 0.0028 rad/s
b) [tex]23.68 m/s^2[/tex]
c) [tex]0 m/s^2[/tex]
Explanation:
a)
When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by
[tex]\omega = \frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement
t is the time interval
The angular speed of an object in circular motion can also be written as
[tex]\omega = \frac{v}{r}[/tex] (1)
where
v is the linear speed of the object
r is the radius of the orbit
For the spaceship in this problem we have:
[tex]v=29,960 km/h[/tex] is the linear speed, converted into m/s,
[tex]v=8322 m/s[/tex]
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Subsituting into eq(1), we find the angular speed of the spaceship:
[tex]\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s[/tex]
b)
When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.
The magnitude of the radial acceleration is given by
[tex]a_r=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
[tex]r[/tex] is the radius of the orbit
For the spaceship in the problem, we have
[tex]\omega=0.0028 rad/s[/tex] is the angular speed
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Substittuing into the equation above, we find the radial acceleration:
[tex]a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2[/tex]
c)
When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.
The tangential acceleration is given by
[tex]a_t=\frac{\Delta v}{\Delta t}[/tex]
where
[tex]\Delta v[/tex] is the change in the linear speed
[tex]\Delta t[/tex] is the time interval
In this problem, the spaceship is moving with constant linear speed equal to
[tex]v=8322 m/s[/tex]
Therefore, its linear speed is not changing, so the change in linear speed is zero:
[tex]\Delta v=0[/tex]
And therefore, the tangential acceleration is zero as well:
[tex]a_t=\frac{0}{\Delta t}=0 m/s^2[/tex]
Suppose an object is located in deep space, far away from any gravity. Would a force need to be applied in order to accelerate the object in space, or would there be no need for a force to make it accelerate?
Answer:
Force will be needed to make the object move
Explanation:
An object at rest will stay at rest and an object in motion will stay in motion unless acted on by an unbalaced object.
Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this problem. Car A uses tires for which the coefficient of static friction is 0.335 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 26.8 m/s. Car B uses tires for which the coefficient of static friction is 0.683 on the same curve. What is the maximum speed at which car B can negotiate the curve?
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
An antenna emits an electromagnetic wave. the electric field lines at a certain instant in time are shown.at point a, what is the direction of the magnetic field?
The direction of the magnetic field at a point in an electromagnetic wave can be determined using the 'right-hand rule'. Point your thumb in the direction of the electric field and curl your fingers. The direction your fingers curl is the direction of the magnetic field.
Explanation:In physics, particularly electromagnetic theory, the electric and magnetic fields of an electromagnetic wave are always perpendicular to each other and to the direction of propagation. The direction of the magnetic field at point 'a' can be determined by using the 'right-hand rule'. This rule states that if you point your thumb in the direction of the electric field and curl your fingers, the direction your fingers curl is the direction of the magnetic field. So, the direction of the magnetic field at point 'a' would be in the direction your fingers curl when you point your thumb in the direction of the electric field at point 'a'.
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What is the period that corresponds to a
frequency of 39.5 Hz?
Answer in units of s.
The answer for the following problem is mentioned below.
Therefore the time period is 0.02 seconds.Explanation:
Frequency:
The number of waves that pass a fixed place in a given amount of time. (or)
The number of waves that pas by per second.
The SI unit of the frequency is Hertz(Hz).
Time period:
The time taken for one complete cycle of vibration to pass a given point.
The SI unit of time period is seconds. (s)
Given:
Frequency (f) = 39.5 Hz
To calculate:
Time period (T)
We know;
According to the problem;
From the problem;
f = [tex]\frac{1}{T}[/tex]
Where;
f represents the frequency
T represents the time period
f = [tex]\frac{1}{39.5}[/tex]
f = 0.02 seconds
Therefore the time period is 0.02 seconds.
Given values,
Frequency,
F = 39.5 HzNow,
→ [tex]Frequency = \frac{1}{Time \ period}[/tex]
or,
→ [tex]F = \frac{1}{t}[/tex]
By substituting the values,
[tex]39.5 = \frac{1}{t}[/tex]
[tex]t = \frac{1}{39.5}[/tex]
[tex]= 0.02 \ seconds[/tex]
Thus the answer above is correct.
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A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck. The puck travels for distance of 0.85 m while experiencing the force from the stick. The puck leaves the stick with a speed of 8.0 m/s and travels at constant speed in a straight line along the horizontal ice for a distance of 12 m. The frictional force between the puck and the ice surface is negligible. How does the magnitude of the force exerted by the stick on the puck Fpuck, stick compare to the magnitude of the force exerted by the puck on the stick Fstick, puck at the time interval in which the stick is in contact with the puck?
A) Fpuck, stick < Fstick, puck
B) Fpuck, stick > Fstick, puck
C) Fpuck, stick = F stick, puck
D) It cannot be determined because the masses of the puck and stick are unknown.
Answer:
D
Explanation:
It cannot be determined because the masses of the puck and stick are unknown.
Answer:
the answer is D
Explanation:
It cannot be determined because the masses of the puck and stick are unknown.
According to the graph, what is the factorization of x2 - 6x + 5?
A. (X - 5)(x - 1)
B. (x + 5)(x - 1)
C. (X - 5)(x + 1)
D. (x + 5)(x + 1)
Answer:
the answer is A (×-5)(×-1)
Answer:
A. (x - 5)(x - 1)
Explanation:
x² - 6x + 5
ax² + by + c
What 2 numbers give a product of 5 (a * c) and a sum of -6 (b)?
-5, -1 ---> -5 + -1 = -6, (-5) * (-1) = 5
x² - 6x + 5 = (x - 5)(x - 1)
Does that help?
Two carts were initially at rest and side-by-side between two photogates. The mass of cart 1 was 0.505 kg. A spring was activated forcing the two carts away from one another, passing through the photogates and producing the readings as shown below. Determine the mass of cart 2.
How does the frequency of gamma rays compare to the frequency of microwaves?
Answer:
Frequency of gamma rays is more than microwave.
Explanation:
The number of vibrations per unit time is called frequency of a wave. The SI unit of frequency is Hertz. It is equal to [tex]s^{-1}[/tex].
The frequency of gamma rays is of the order of [tex]10^{20}\ Hz[/tex]. The frequency of microwave is of the order of [tex]10^8\ Hz[/tex].
It is clear that gamma rays have more frequency that of the microwave.
ck-12 Refraction
2/10
SKILL LEVEL
To be de
Light coming straight from a laser pointer passes from air into a
tub of liquid. It hits the surface of the liquid at an angle of 28°
from vertical, and continues into the liquid at an angle of 22°.
What is the index of refraction of the substance in the tub? n =
Answer:
1.25
Explanation:
From refractive index of the substance in the tub, n = sini/sinr where i = angle of incidence = 28° and r = angle of refraction = 22°
So, n = sini/sinr = sin28°/sin22° = 1.253 ≅ 1.25
The index of refraction of the substance in this tub is equal to 1.25.
Given the following data:
Angle of incidence = 28°.Angle of refraction = 22°.What is Snell's law?Snell's law gives the relationship between the angle of incidence and angle of refraction with respect to light or other waves passing through a media or two different substances such as glass, water or air.
Snell's Law states that the when light travels from one medium to another, it generally refracts. Mathematically, it is given by this formula:
[tex]n=\frac{sini}{sinr}[/tex]
Substituting the given parameters into the formula, we have;
[tex]n=\frac{sin28}{sin22}\\\\n=\frac{0.4695}{0.3746}[/tex]
n = 1.25
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The diagram shows the mechanical parts of scissors
Answer: B. Both involve the use of levers.
Explanation:
You need to design a photodetector that can respond to the entire range of visible light. True or False
Answer: True
Explanation:
A photo detector that can respond to the entire rang of visible light can be design, it is true.
Photo detector is a device in an optical receiver which receives optical signals and convert it to electric signal. It is the key device position in front of the optical receiver.
The upward force exerted on an object falling through air is _____.
A. terminal velocity
B. Momentum
C. Air Resistance
D. Weightless
(C) Air Resistance
Explanation:
When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.
The upward force exerted on an object falling through air is ______.
A. terminal velocity B. momentum
C. air resistance D. weightless
The upward force exerted on an object falling through air is air resistance. The correct option is C.
What is air resistance?When an object moves through air, it creates a force called air resistance. This force basically acts in the inverse way of direction of a body moving via air.
The frictional force of air resistance acts on the moving body. When a body moves, air resistance slows it down.
When an object moves through the air, it encounters air resistance. Resistance varies depending on the velocity, shape, and area of the object. The higher the air resistance, the faster an object moves and the larger its area.
When air pushes against a moving object, it creates air resistance force. Frictional forces include air resistance. Force is always applied against the motion of an object.
Thus, the correct option is C.
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A calorimeter contains 0.50 kg of water at 15°C. A 0.040-kg block of zinc at 115°C is placed in the water. The specific heat of zinc is 388 J/kg °C and the specific heat of water is 4184 J/kg °C
What is the final temperature of the system?
Answer:
The final temperature is 16°C
Explanation:
mcΔT (water) = mcΔT (zinc)
0.50(4184)(t-15) = 0.040(388)(115-t)
2092(t-15) = 15.52(115-t)
2092t-31380 = 1784.8-15.52t
2029t+15.52t= 1784.8+31380
2044.52t = 33164.8
t = 33164.8/2044.52
t = 16
You perform an experiment with a long column of air and a tuning fork. The column of air is defined by a very long vertical plastic tube with a circular cross section; the upper end of the tube is open to the outside air and the lower end of the tube is filled with water. The column of air extends from the top of the water to the open end of the tube; the length of the column can be varied by changing the water level in the lower end of the tube. (The columns of air in this problem are longer than those you will have used in lab; this allows us to imagine the use of lower frequency tuning forks than those you will have used in lab.)
ou are using a tuning fork of frequency 256 Hz. For one particular water level, you hear a loud sound when the fork is struck and positioned just above the open end of the tube. When the water level is lowered by 676 mm, you again hear a loud sound.
(a) What is the wavelength of the 256 Hz sound waves in the column of air? Give your answer in meters, not mm. m
(b) What is the speed of sound in the column of air? m/s
Answer:
[tex]\lambda=4L=1.33m[/tex]
v=343m/s
Explanation:
We have to take into account the expressions
[tex]f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}[/tex]
if we assume that 256Hz is the fundamental frequency we have
[tex]f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m[/tex]
and for wavelength
[tex]\lambda=4L=1.33m[/tex]
hope this helps!!
Answer:
a) The wavelength, λ = 1.352 m
b) The speed of the sound in the column air is v = 346.112 m/s
Explanation:
a) Frequency, f = 256 Hz
The water level is lowered by 676 mm. i.e. x = 676 * 10⁻³ m
The length, x, at which the loudest sound will be heard is λ/2
i.e. x = λ/2
Therefore the wavelength of the 256 Hz sound waves in the column air is given by the relation
λ = 2x
λ = 2 * 676 * 10⁻³
λ = 1.352 m
b) The speed of the sound in the column air:
The speed, v, is given by the relation v = f λ
speed, v = 256 * 1.352
v = 346.112 m/s
ptsA flashlight is pointed at the ground, forming a spot of light. The radius of the spot of light is measured and the area of the spot is calculated. The brightness of the spot is measured. Then the flashlight is moved so it is twice as far from the floor.The new brightness will be ________ times the old brightness.
Answer:
I₂ = I₀ 1/4
Explanation:
For this exercise we use the definition of intensity which is the power per unit area
I = P / A
The emitted power is constant, so
P = I A
We can write this equation for the start and end point with index 2
I₀ A₀ = I₂ A₂
I₂ = I₀ A₀ / A₂
The spot area is the area of the circle
A₀ = π r₀²
We substitute
I₂ = I₀ r₀² / r₂²
It indicates that the radius of the spot is twice the initial radius
r₂ = 2 r₀
I₂ = I₀ (r₀ / 2 r₀)²
I₂ = I₀ 1/4
The position of a full moon is located
Answer:
opposite the sun. between the Earth and the sun. rising perpendicular to the sun.
Explanation:
Two balls of the same shape and size have charges of +8 and-2. The balls are brought together, allowed to touch, and then separated. What is the net charge on each ball now?
Explanation:
Initially, two balls of the same shape and size have charges of +8 and-2. The balls are brought together, allowed to touch, and then separated.
When they are touched, the total charge is evenly distributed among the whole surface and they behaves as if it is a single piece. Net charge at this point is +8-2=6 C. When they are separated, each ball will have a charge of 6/2 = 3 C.
When two identical balls with charges of +8 and -2 are brought into contact and then separated, each ball ends up with a net charge of +3, due to the conservation of charge and even distribution across identical conducting surfaces.
When two balls of the same shape and size with charges of +8 and -2 are brought together, allowed to touch, and then separated, the net charge on each ball is a result of the equal distribution of the total charge across both balls. Initially, the total charge is the sum of +8 and -2, which equals +6. Since the balls are identical and they come into contact, this total charge of +6 is evenly divided between them. Therefore, after they are separated, each ball will have a net charge of +3.
This occurs because, when conducting objects touch, charges redistribute themselves evenly across the surfaces of the objects until they reach an equilibrium while maintaining the conservation of charge. In simpler terms, the total quantity of charge is conserved, and since the two balls are identical, the charge divides evenly.
How does an unbalanced force affect an object?
Answer:
it can change the speed only or can change direction only or can change both
Imagine two circular plates; one is solid and the other has a hole cut out of the center. Both plates have the same radius, same thickness, and same mass. The same force F is applied tangential to the edge of each plate in such a way that the plates rotate about an axis passing through the center and perpendicular to the surface of the plates.
Which of the following statements helps to explain the question asked above? (Select all that apply.)
Both plates will be subjected to the same torque.
Angular acceleration is inversely proportional to the moment of inertia.
Angular acceleration is directly proportional to the moment of inertia.
The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.
Because both plates have the same mass, they will have the same moment of inertia.
Answer:
Both plates will be subjected to the same torque. TRUE Angular acceleration is inversely proportional to the moment of inertia. TRUE The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia. TRUEExplanation:
Both plates will be subjected to the same torque. TRUE Angular acceleration is inversely proportional to the moment of inertia. TRUE The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia. TRUEWe have that for the Question "Imagine two circular plates; one is solid and the other has a hole cut out of the center. Both plates have the same radius, same thickness, and same mass. The same force F is applied tangential to the edge of each plate in such a way that the plates rotate about an axis passing through the center and perpendicular to the surface of the plates."
Question: Which one of the following statements is true regarding the angular acceleration?
Answer : The solid plate will have the greater angular acceleration
Question : Which of the following statements helps to explain the question asked above? (Select all that apply.)
Answer :
(a) Both plates will be subjected to the same torque.(b) Angular acceleration is inversely proportional to the moment of inertia. (d) The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.For more information on this visit
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A person is on an elevator that moves downward at 1.8 m/s². If the person weighs 686 N, what is the net force on the person?
Answer:
Explanation:
Hi
The net force on a person in a descending elevator can be calculated using Newton's second law of motion, considering the person's weight and the elevator's downward acceleration.
The problem relates to Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In the context of a person in a descending elevator, the net force exerted on the person by the floor, sometimes referred to as the normal force, can be found by considering both the gravitational force (weight) acting downwards and the acceleration of the elevator.
Given the person's weight (W) is 686 N and the downward acceleration (a) of the elevator is 1.8 m/s², we can first find the person's mass (m) by rearranging the formula for weight (W = mg), where g is the acceleration due to gravity (9.8 m/s²), giving us m = W/g. Secondly, we can calculate the net force (Fnet) using Newton's second law, taking into account that the actual acceleration of the person is the difference between the acceleration due to gravity and the elevator's acceleration since the elevator is moving downward. Therefore, Fnet = m(g - a).
Net force acting on a person can be calculated as the difference between the force of gravity acting on the person and the force exerted by the elevator. First, calculate the force of gravity: Fg = mg, where m is the mass and g is acceleration due to gravity (9.8 m/s²). Then, calculate the force exerted by the elevator using the equation Fnet = ma, where a is the acceleration of the elevator. Finally, find the net force by subtracting the force of gravity from the force exerted by the elevator.
The wavelength of a sound wave in this room is 1.13 m and the frequency is 301 Hz.
a. What is the speed of the wave in the room?
©Modeling Instruction Program 2004 3 W2, Mechanical Waves in 1D, WS 5 v3.2
b. If you double the frequency of the sound wave, determine its speed.
c. What happens to the wavelength if you cut the frequency in half? How do you know?
I'll throw in brainiest too.
Answer:
a. 340.13 m/s b. 680.26 m/s c. our wavelength doubles
Explanation:
a. speed of wave, v = fλ were f = frequency = 301 Hz and λ = wavelength = 1.13 m.
v = fλ = 301 Hz × 1.13 m = 340.13 m/s
b. If we double the frequency then f = 2 × 301 Hz = 602 Hz
v = fλ = 602 Hz × 1.13 m = 680.26 m/s
c. If the speed of the wave is still 340.13 m/s, if we cut the frequency in half, then frequency now equals f = 301 Hz/2 = 150.5 Hz.
Since v = fλ,
λ = v/f = 340.13 m/s ÷ 150.5 Hz = 2.26 m.
Since our initial wavelength λ₀ = 1.13 m,
λ/λ₀ = 2.26 m/1.13 m = 2.
So, λ = 2λ₀ our wavelength doubles
A 7.80-g bullet moving at 540 m/s strikes the hand of a superhero, causing the hand to move 5.10 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet.
Answer:
[tex]F = 22298.824\,N[/tex]
Explanation:
According to the Principle of Energy Conservation and the Work-Energy Theorem, the bullet has the following expression:
[tex]U_{g,A} + K_{A} = U_{g,B} + K_{B} + W_{loss}[/tex]
[tex]W_{loss} = U_{g,A}-U_{g,B} + K_{A}-K_{B}[/tex]
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot [v_{A}^{2}-v_{B}^{2}][/tex]
The average force exerted on the bullet to stop it is:
[tex]F = \frac{m\cdot [v_{A}^{2}-v_{B}^{2}]}{2\cdot \Delta s}[/tex]
[tex]F = \frac{(7.8\times 10^{-3}\,kg)\cdot [(540\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}]}{2\cdot (0.051\,m)}[/tex]
[tex]F = 22298.824\,N[/tex]
Answer:
22298.82N
Explanation:
The bullet has a kinetic energy: ½*m*v²
mass of bullet = 7.80g = 7.80÷1000 = 0.0078kg
distance in meter =5.10cm = 5.10÷100 = 0.051meter
K.e = 0.5 ×0.0078 × 540× 540
K.e = 1137.24 J
When the bullet stops this energy goes to zero, as the energy must be conserved the work done by the head of the superhero must be equal to the original energy of the bullet:
W= K
Considering an average constant force, the work can be calculated as:
W=F*d=K
Solving for F:
F = K/d
1137.24 J/ 0.051m
= 22298.82N
a planet moves around the sun is nearly circular orbit its period of revolution t depends upon
1)radius r of orbit.
2)mass m of the sun.
3) gravitational contact. show dimensionally t^2 proportional to
[tex] {r}^{3} [/tex]
Answer:
t² ∝ r³
Explanation:
Let m be the mass of the sun and m₁ be the mass of the planet and r its distance from the sun. Its gravitational force of attraction equals its centripetal force. So,
Gmm₁/r² = m₁rω² = m₁r(2π/t)² were t is the period of orbit of the planet
Gm/r² = 4π²r/t²
rearranging we ave
t² = (4π²/Gm)r³ since k = 4π²/Gm = constant
t² = kr³ and
t² ∝ r³
What’s the formula for work
Answer:
Fd
Explanation:
Work is force times distance. If you push on an object really hard but it does not budge, you have still performed no work on it, because anything times zero is still zero.
Answer:
W = F * d
Explanation:
9. (True or False) During an early morning rain shower, Jim
could see a rainbow to the north as the sun's rays in the east
broke through the clouds, illuminating the landscape.
The diagram shows a person using a piece of gym equipment to lift weights.
A man sits on a chair with pulleys and a line attached to the side. His leg has the line attached to it and is shown in solid lines with his foot on the floor. His leg is also shown raised with a dotted extension of the line attached to it.
Which best describes the body in terms of simple machines?
The knees are inclined planes, and the lower legs hold the loads. The ankles are the fulcrums.
The lower legs are inclined planes, and the knees hold the loads. The ankles are the fulcrums.
The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.
The knees are levers, and the lower legs are fulcrums. The ankles hold the loads.
Final answer:
In the provided scenario, the correct description of the body in terms of simple machines is that the lower legs are levers and the knees serve as fulcrums.
Explanation:
When we analyze the biomechanics of the body using the piece of gym equipment in the provided scenario, particularly during a leg workout, we can compare the body to a set of simple machines. In this context, the lower legs act as levers.
The knees operate as fulcrums, which are the points about which the levers rotate. The ankles do not carry or hold the load in this action; rather they act as the point of application for the effort when extending the foot in movements like calf raises.
The configuration described is analogous to a first-class lever, where the fulcrum (knee) is situated between the effort (muscles applying force near the ankle) and the load (the resistance on the leg extension machine). Hence, the correct description of the body in terms of simple machines is: "The lower legs are levers, and the knees are fulcrums."
To address the components from the question:
The knees are the fulcrums since they are the pivot point for the motion of the lower leg.
The lower legs are levers because they move around the fulcrum to lift the load on the gym equipment.
The ankles are where the calf muscles apply the effort but are not where the load is held.
Therefore, third option is correct: The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.
Nuclear power plants are designed to convert nuclear energy into what type of energy
Answer:
Nuclear plants are plants that involve the fission of relatively large atoms to produce electricity as its final product.
In nuclear power plants, the nuclear energy is converted to heat energy, after which the heat energy is converted to mechanical( rotational energy) which helps in the spinning of the turbines. The mechanical energy is then converted to electrical energy.
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.05 m 1.05 m long rod as you jog at 3.27 m/s 3.27 m/s , holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.275 mV 0.275 mV across the rod? Express the answer in milliteslas.
Answer:
The strength of the magnetic field is 0.08 mT
Explanation:
Given:
Length of rod [tex]l = 1.05[/tex] m
Velocity of rod [tex]v = 3.27[/tex] [tex]\frac{m}{s}[/tex]
Induced emf [tex]\epsilon = 0.275 \times 10^{-3}[/tex] V
According to the faraday's law
We know that the induced emf of rod is given by,
[tex]\epsilon = Blv[/tex]
Where [tex]B =[/tex] magnetic field
For finding the magnetic field,
[tex]B = \frac{\epsilon }{lv}[/tex]
[tex]B = \frac{0.275 \times 10^{-3} }{1.05 \times 3.27}[/tex]
[tex]B = 0.08 \times 10^{-3}[/tex]
[tex]B = 0.08[/tex] mT
Therefore, the strength of the magnetic field is 0.08 mT
Look at the diagram showing resistance and flow of electrons. A top box labeled X contains 2 circles with plus signs and 2 circles with minus signs. A bottom box labeled Y contains 4 circles with minus signs and 8 circles with plus signs. An arrow Z runs from the bottom box to the top box. Which labels best complete the diagram? X: High resistance Y: Low resistance Z: Flow of electrons X: Low resistance Y: Flow of electrons Z: High resistance X: Flow of electrons Y: High potential energy Z: Low potential energy X: Low potential energy Y: High potential energy Z: Flow of electrons
Answer:
D: X: Low potential energy
Y: High potential energy
Z: Flow of electrons
Explanation: trust
Answer:
d
Explanation: