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).
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
what is the definition of a moment of force?
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
Write about
a time you had to ride a bicycle on a difficult
surface. What did you have to do to adjust your
riding?
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
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
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
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?
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.
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?
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
While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?
Answer:
-2.83 m/s²
Explanation:
Initial velocity (u) = 34 m/sFinal velocity (v) = 17 m/sTime taken (t) = 6 seconds❖ Acceleration is defined as the rate of change in velocity with time.
→ a = (v - u)/t
v denotes final velocitya denotes accelerationu denotes initial velocityt denotes time→ a = (17 - 34)/6 m/s²
→ a = -17/6 m/s²
→ Acceleration = -2.83 m/s²(Minus sign implies that the velocity is decreasing.)
If a Sam goes above and beyond what is expected in her job duties, what might Sam's supervisor note on Sam's next evaluation?
O A. Sam has good attendance.
O B. Sam shows initiative.
O C. Sam stays focused on assigned tasks.
OD. Sam works well with others.
Answer:
2
Exadadadwwwwwwwplanation:
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?
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.
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
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
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
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]
Which planet has the greatest gravity (of the choices shown)
Saturn
Venus
Uranus
Earth
Answer:
Earth
Explanation:
Answer:
Saturn/has best gravity out of the 4 planets.
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?
Answer:
hope u can understand the method
PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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)
The best way to prepare for a workout is to stretch key muscles first and follow the stretching with a jog to get the blood flowing to the muscles.
Explanation:
stretching out and daily exercise
Answer:
false
Explanation:
In designing subway trains and stations, engineers want to optimize the time and energy use between stations. A train can comfortably accelerate at 5 ft/s2 without causing its passengers to fall over. When coasting, the train will slow down at an acceleration of -2 ft/s2. The train begins at rest and must come to a complete stop at the destination. If two train stations are 2 miles apart, determine the fastest time that the train can arrive at the second station given the provided accelerations. Also, determine the maximum velocity obtained. Draw the a-t, v-t, and s-tgraphs for the train. (122 s, 174 ft/s)
Answer:
Explanation:
Let the time required for acceleration a₁ and deceleration a₂ be t₁ and t₂ .
Since final velocity during acceleration and initial velocity during deceleration are same
a₁ t₁ = a₂ t₂
5t₁ = 2 t₂ ------------------------------------------ ( 1 )
Distance travelled during acceleration = 1/2 a₁t₁²
= 1/2 x 5 x t₁² = 2.5 t₁²
Distance travelled during deceleration = 1/2 a₂t₂²
= 1/2 x 2 x t₂² = t₂²
Total distance travelled = 2 miles = 2 x 1760 x 3 ft = 10560
2.5 t₁² + t₂² = 10560
2.5 ( 2t₂ / 5 )² + t₂² = 10560
.4 t₂² + t₂² = 10560
1.4 t₂² = 10560
t₂ = 86.85 s
t₁ = 2t₂ / 5 = 34.75 s
t₁ + t₂ = 121.6 = 122 s
Total time taken = 122 s .
maximum velocity = a₁t₁
= 5 x 34.75 = 173.75 = 174 m/s .
Which of the following would cause electromagnetic induction?
The current in a loop of coils is constant.
A loop of coils is placed near high voltage power lines with alternating current.
A constant current is supplied to a circuit board in your computer.
Your phone is turned on after being fully charged.
Answer:
The loop of coils - Electromagnetic induction is caused a changing magnetic field moving thru a closed loop(s) of coils
Please help if you can!
Answer:
C, Red has the longest one
c red
red has longest wavelength
amnh dot org
Which of the following should be practiced among all co-workers in the workplace?
A. bullying
B. discrimination
C. stereotyping
D. tolerance

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
Answer:
12558 J
Explanation:
Please do mark as brainliest. Hope this helps! :)
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.
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%
What step occurs first in a scientific investigation and in the development of a new technology?
-Findings are communicated to others.
-A problem or need is identified
-A series of tests are analyzed.
-Information is researched
Answer:
B. A problem or need is identified
Explanation:
hoep this helps
Answer:
B
Explanation:
edge
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
Answer:
the answer of this question is true
What does it mean when work is positive?
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
differences between adhesion and cohesion
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.
True or false. When a girl walks the action of pushing and the equal amd opposite reaction is being projected forward
This is true I think
It applies to Newton's Laws
it's true because it's a part of newtons law
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
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
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.
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.