A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B field (B is constant in time) and then back into a field free region to the left. The self inductance of the loop is negligible. True While the loop is entirely in the field, the emf in the loop is zero. False When entering the field the coil experiences a magnetic force to the left. False When leaving the field the coil experiences a magnetic force to the right. True Upon entering the field, a clockwise current flows in the loop.
Answers
Answer:
False, True, False, True
Explanation:
Given a square loop of a wire having a small resistance and moving with a constant sped from the free region to a region having a region having a uniform field B. It is then moved back to the left in the free field region.
For the following statements :
While the loop is entirely in the field, the emf in the loop is zero. -- TRUE
The emf of the coil is zero when the coil moves into the field from a free region field to a uniform field region field.
When entering the field the coil experiences a magnetic force to the left ---- FALSE
According to the Lenz law, a loop entering the field opposes it and moves towards the right.
When leaving the field the coil experiences a magnetic force to the right. --- TRUE
Upon entering the field, a clockwise current flows in the loop. ---- TRUE
In the field the coil or the loop tries to oppose the magnetic field and forces the current in the loop to move in a clockwise direction.
Related Questions
Please help if you can!
Answers
Answer:
C, Red has the longest one
c red
red has longest wavelength
amnh dot org
How much heat is required to raise the temperature of 50 grams of water from 30 °C to 90 °C? C of water 4186 J / kg C.
12558 J
12558000 J
125580 J
1255800 J
Answers
Answer:
12558 J
Explanation:
Please do mark as brainliest. Hope this helps! :)
A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answers
Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the backpack.
Answers
Answer:
2420 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 22.9 N
Angle (θ) = 35°
Distance (d) = 129 m
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 22.9 × 129 × Cos 35
Wd = 22.9 × 129 × 0.8192
Wd ≈ 2420 J
Thus, the workdone is 2420 J.
differences between adhesion and cohesion
Answers
Answer:
As for the definitions, the tendency of two or more different molecules to bond with each other is known as Adhesion, whereas the force of attraction between the same molecules is known as Cohesion.
hopefully this helps
Answer:
Adhesion is the force of attraction between molecules of different substances while cohesion is the force of attraction between molecules of same substances.
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is
Answers
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency [tex]\zeta[/tex] = 3%
where;
[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]
[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]
[tex]Q_{supplied } = 66.66 \ MW[/tex]
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]
Also;
[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]
[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]
[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]
[tex]LMTD = \dfrac{8}{In (5)}[/tex]
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
[tex]Q_H = UA (LMTD)[/tex]
where;
U = overall heat coefficient given as 1200 W/m².K
[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]
The mass flow rate:
[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]
what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of 2.7m/s and air resistance is negligible
Answers
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision
Answers
Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
[tex](2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s[/tex]
Now, we will apply the law of conservation of momentum:
[tex]m_1v_1 = m_2v_2[/tex]
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
[tex](3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s[/tex]
Now, we again use the third equation of motion for the upward motion of the ball:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
[tex](2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\[/tex]
h = 3.5 m
On a distance-time graph, what is shown when the curve is flat going from left to the right?
A. a negative speed
B. no speed
C. a positive speed
D. It does not mean anything.
Please help me !!im on a test
Answers
What does it mean when work is positive?
Answers
Answer:
When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work.
Explanation:
Hope this helps you
When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work
So yeah basically what the person above me said
what is the definition of a moment of force?
Answers
Answer:
The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis.
Answer:
Torque
Explanation:
I serached it up sjdjbdjd
A 10kg block is Pulled along a horizontal
Surface by a force
of 50N at an angles
of 37° with the horizontal If the
coefficient of sliding friction b/n the
block and the surface is o.2
(g=10m/s^2 Sin 37=O.6 and cos 37 = 0.8)
A, what frictional forces acting on the block?
B,what is the acceleration of the block?
Answers
Answer:
hope u can understand the method
Electroconvulsive therapy would be done under the
supervision of a counseling psychologist, where high level
of electric shock would be admistered.
Select one:
True
False
Answers
Answer:
the answer of this question is true
A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?
Answers
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]
a = 2.44 m/s²
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]
s = 63.24 m
Se transmiten ondas transversales en una cuerda tensada orientada sobre el eje x. La función de ondas correspondientes es y= 5.00 sen(6.05x + 5.19t + 1.57). Donde y y x están en metros y t en segundos. Cual es la rapidez de las ondas que se transmiten en dicha cuerda?
Answers
Answer:
Explanation:
Whats the entire question?
An object of mass 45 kg is observed to accelerate at the rate of 6 m/s2. Calculate the force required to produce this acceleration
Answers
The density of table sugar is 1.59g/cm3 what is the volume of 7.85g of sugar?
Answers
Answer: 4.94cm³
Explanation:
Data;
ρ = 1.59g/cm³
mass = 7.85g
volume = ?
density = mass / volume
ρ = m / v
v = m / ρ
v = 7.85 / 1.59
v = 4.94cm³
An 800 kg charging bull rams through a wooden fence. It was travelling at
5 m/s, now it's travelling at 3 m/s. How much impulse did the bull
experience by smashing the fence?
Answers
Answer:
J = 1600 kg-m/s
Explanation:
Given that,
The mass of charging bull rams, m = 800 kg
Initial speed, u = 5 m/s
Final speed, v = 3 m/s
We need to find the impulse the bull experience by smashing the fence. Let it is J. We know that, impulse is equal to the change in momentum such that,
J = m(v-u)
Put all the values,
J = 800(3-5)
= 800(-2)
= -1600 kg-m/s
Hence, the magnitude of impulse is equal to 1600 kg-m/s.
A man applies a force of 315 N to push the block of 225 kg onto 10 m ramp. Calculate the efficiency of the person if the mass gains a height of 1.2 m.
Answers
Answer:
η = 0.84 = 84%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the block = mgh
input = work done by man = Fd
Therefore,
[tex]\eta = \frac{mgh}{Fd}[/tex]
where,
m = mass of block = 225 kg
g = acceleration due to gravity = 9.81 m/s²
h = height gained by block = 1.2 m
F = force exerted by man = 315 N
d = distance covered by man = 10 m
Therefore,
[tex]\eta = \frac{(225\ kg)(9.81\ m/s^2)(1.2\ m)}{(315\ N)(10\ m)}[/tex]
η = 0.84 = 84%
Calculate the potential difference across the 8 ohm resistor
Answers
Explanation:
if the current is 1A
V=iR
V= 1 × 8
V = 8volts
PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answers
Answer:
A) pass the ball to a teammate
B) Smash the shuttlecock downward in your opponents court
C)Do a fake hit . . .
D) Do a fake hit . . .
(Best guess)
g ou drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision
Answers
Answer:
0.4112 m
Explanation:
The mass of the 1st ball = 3.6 kg
The height of the 1st ball =3.5 m
The mass of the 2nd ball = 3.6 kg
Mass of the bar M = 9.9 kg
Length of the bar L = 4.2 m
The velocity of the ball when it dropped from the height is calculated by using the formula:
[tex]\dfrac{1}{2}mv_1^2 = mgh_1 \\ \\ v = \sqrt{2gh_1} \\ \\ v =\sqrt{2\times9.8 \times 3.5} \\ \\ v = 8.283 \ m/s\\\\[/tex]
Provided that the bar is pivoted at the center and the ball is placed at the two ends, the moment of inertia for the bar is:
[tex]I = \dfrac{1}{12}ML^2 + m_1 (\dfrac{L}{2})^2 + m_2(\dfrac{L}{2})^2 \\ \\ =\dfrac{1}{12}(9.9kg)(4.2m)^2 + [3.6 kg+3.6kg](\dfrac{4.2}{2 \ m})^2 \\ \\ = 46.305 \ kg.m^2[/tex]
The angular momentum of the system due to the ball can be determined by using the formula:
L = mvr
L = (3.6 kg) (8.283 m/s) (2.1 m)
L = 62.61948 kg. m²
Now, Using the law of conservation:
[tex]L_i = L_f \\ \\ 62.61948 \ kg.m^2/s = I \omega \\ \\[/tex]
[tex]\omega = \dfrac{62.6198 \ kg.m^2/s}{46.305 \ kg.m^2}[/tex]
[tex]\omega =1.352 \ rad/s[/tex]
The linear angular velocity is deduced to be:
[tex]v = r \omega \\ \\ v = (2.1 \ m) ( 1.352 \ rad/s) = 2.839 \ m/s[/tex]
∴
the height raised by the second ball is:
[tex]h_2 = \dfrac{v^2}{2g} \\ \\ h_2 = \dfrac{(2.839)^2}{2(9.8 \ m/s^2)} \\ \\ h_2 =0.4112 \ m[/tex]
True or false. When a girl walks the action of pushing and the equal amd opposite reaction is being projected forward
Answers
This is true I think
It applies to Newton's Laws
it's true because it's a part of newtons law
When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string
Answers
Solution :
Mass is varied keeping frequency constant.
Wavelength, λ [tex]$=\frac{2l}{n}$[/tex]
where length of spring = l
number of segments = n
Velocity, v = λ x f
= [tex]$\sqrt{\frac{T}{\mu}}$[/tex]
[tex]$\mu $[/tex] = mass density, T = tension in string
[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]
[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex] , n = 2
[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]
[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]
[tex]$\Rightarrow m = 2.619\ kg$[/tex]
Therefore, μ = 0.002785 kg/ m
Frequency is varied keeping T constant
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]
[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]
f = 210 Hz
when two capacitor 3muF and 6muF are connected in a parallel and combination is charged to a potential of 120 volt the potential difference across the 3muF capacitor is
Answers
Answer:
V₁ = V = 120 V
Explanation:
Such a combination of capacitors in which;
1- Potential difference across each capacitor is the same
2- Total charge is distributed amongst the capacitors
; is called Parallel Combination.
Therefore, in this case, the potential difference across each capacitor will also be the same. Because the capacitors are connected in parallel here. So the voltage across 3 μF capacitor will be the same as the voltage across the 6 μF capacitor and they both will be equal to the total potential difference.
V₁ = V = 120 V
Which is the best way to become familiar with your company's policies and procedures?
O
A. ask the person who hired you
O
B. look in the employee handbook
C. tell your supervisor you need help
D. visit the company's website
Answers
The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically accessible quantum state of a system has equal probability of being populated, which in turn leads to the Boltzmann distribution for a system in thermal equilibrium.
a. True
b. False
Answers
Answer:
Hello! Your answer would be, A) True
Explanation:
Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥
A 25.0 kg mass is traveling to the right with a speed of 2.80 m/s on a smooth horizontal surface when it collides with and sticks to a second 25.0 kg mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant 170.0 N/m. The other end of the spring is fixed to a wall to the right of the second mass.
1) Find the frequency of the subsequent oscillations.2) Find the amplitude of the subsequent oscillations.3) Find the period of the subsequent oscillations.4) How long does it take the system to return the first time to the position it had immediately after the collision?
Answers
Answer:
Explanation:
1 ) angular frequency ω = √ ( k / m )
=√ ( 170 / 50 )
= 1.844 rad /s
2πn = 1.844 where n is frequency of oscillation
n = 1.844 / (2 x 3.14 )
= .294 per sec
= .294 x 60 = 18 approx. per minute .
Velocity just after collision of composite mass ( using law of conservation of momentum )
= 25 x 2.8 / 50
v = 1.4 m/s
If new amplitude be A
1/2 k A² = 1/2 m v²
m = 25 + 25 = 50 kg
170 x A² = 50 x 1.4²
A = 0.76 m
3 ) period of oscillation = 1 /n
= 1 / .294
= 3.4 s
4 ) It will take complete one period of oscillation ie 3.4 s to come to its original position.
Two parallel copper rods supply power to a high-energy experiment, carrying the same current in opposite directions. The rods are held 8.0 cm apart by insulating blocks mounted every 1.5 m. If each block can tolerate a maximum tension force of 200 N, what is the maximum allowable current
Answers
Answer:
the maximum allowable current is 7302.967 amperl
Explanation:
The computation of the maximum allowable current is shown below;
Force F = mean ÷ 4π 2 I_1 I_2 ÷d × ΔL
200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5
200 = 3.75 × 10^-6 I^2
I = √200 ÷ √ 3.75 × 10^-6
= 7302.967 amperl
Hence, the maximum allowable current is 7302.967 amperl
Basically we applied the above formula
Write about
a time you had to ride a bicycle on a difficult
surface. What did you have to do to adjust your
riding?