Answer:
The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.
Explanation:
A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.
Let us find the drag force which is proportional to the velocity of two conducting plates.
The EMF that is induced in the eddy currents are : [tex]$E=v(B \times L)$[/tex]
The force which is due to the induced magnetic field is, [tex]$F=l(L \times B)$[/tex]
Therefore, [tex]$F=\frac{E}{R} \times (L \times B)$[/tex]
[tex]$F=\frac{v(B \times L)}{R} \times (L \times B)$[/tex]
Here, force is directly proportional to the velocity of the two conducting plates.
Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.
. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder
Answer:
As we can see, a string is attached with block A, and three string is folded with ply which is attached with B
x
B
=3x
A
Now differentiate with respect to x
V
B
=3V
A
Given,
V
A
=0.6m/s(totheright)
So,
V
B
=0.6×3
=1.8m/s(downward)
Explanation:
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Define threshing?
Brainliest For the right answer
Answer:
Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.
Answer:
Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.
The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.
In old times, grain had to be beaten out of the grass on a Threshing Floor.
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and forth between point A, where the string is at the maximum angle of 35.0∘ to the left of vertical, and point C, where the string is at the maximum angle of 35.0∘ to the right of vertical. The string is vertical when the sphere is at point B. Calculate how much work the force of gravity does on the sphere from A to B.
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer
at the bottom of the slide.
Answer:
the velocity of the swimmer at the bottom of the slide is 6.26 m/s
Explanation:
The computation of the velocity of the swimmer at the bottom of the slide is given below:
v = √2gh
= √2 × 9.8 × 2
= 6.26 m/s
Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s
The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change with reference to time is referred to as the velocity
What is velocity?The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
The velocity is found as;
[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]
Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.
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When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)
Answer:
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]
Explanation:
This is an interesting problem, no data is given, so the result is a general expression.
Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂
we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved
initial instant. Just before the crash
L₀ = L₁ + L₂
with
L₁ = I₁ w₁
the moment of inertia of a disc with an axis passing through its center is
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂
final instant. Right after the crash
L_f = I w
in angular momentum it is a scalar quantity, so it is additive
I = I₁ + I₂
angular momentum is conserved
L₀ = L_f
I₁ w₁ + I₂ w₂ = I w
w = [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex] (1)
We already have the angular velocities of the system, let's find the kinetic energy of it
initial
K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²
final
K_f = K = ½ I w²
the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses
ΔK = K_f -K₀
ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²) (2)
In this chaos we know all the values for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated
Now if all this variation of energy turns into heat
Q = ΔK
m_{total} c_e ΔT = ΔK
where the specific heat of the bear discs must be known, suppose they are of the same material
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex] (3)
to make a special case, we suppose some data
the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC
we look for the angular velocity
I₁ = I₂ = I₀
I = 2 I₀
we substitute in 1
w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io
w = w₁ /2
we look for the variation of the kinetic energy with 2
ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)
ΔK = ¼ I₀ w₁² -½ I₀ w₁²
ΔK = - ¼ I₀ w₁²
the negative sign indicates that the kinetic energy decreases
We look for the change in Temperature with the expression 3
ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce
ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]
ΔT = [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]
ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]
in this expression all the terms are contained
The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
What is internal energy?The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.
The given data in the problem is;
[tex]\rm \omega_1[/tex] is the angular velocity of disk 1
[tex]\rm \omega_2[/tex] is the angular velocity of disk 2
r₁ is the radius of disk 1
r₂ is the radius of disk 2
m₁ is the mass of disk 1
m₂ is the mass of disk 2
Momentum before the collision;
[tex]\rm L_1 = I_1 \omega_1[/tex]
The moment of inertia of disc 1
[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]
The momentum gets conserved;
[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]
The change in the kinetic energy is;
[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]
The change in the energy gets converted into heat;
[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]
The change in the temperature is
[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]
The internal energy change is found by;
[tex]\rm \triangle E = mc_v dt[/tex]
[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]
Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 734.1 rad/s 734.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds0.569 seconds , what is the magnitude of the average angular acceleration of the disk
Answer:
[tex]1290.16\ \text{rad/s}^2[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 734.1 rad/s
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time = 0.569 seconds
[tex]\alpha[/tex] = Angular acceleration
From the kinematic equations of rotational motion we have
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2[/tex]
The magnitude of the average angular acceleration of the disk is [tex]1290.16\ \text{rad/s}^2[/tex].
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:[tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]
Now, we need first to find the value of the angular acceleration, that we can get from the following expression:[tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]
Since the machine starts from rest, ω₀ = 0.We know the value of ωf₁ (the operating speed) in rev/min.Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:[tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]
Replacing by the givens in (2):[tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]
Solving for α:[tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]
Replacing (5) and Δt in (1), we get:[tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]
in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:[tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]
In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:[tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]
First of all, we need to find the value of the angular acceleration during the second period.We can use again (2) replacing by the givens:ωf =0 (the machine finally comes to an stop) ω₀ = ωf₁ = 57.5 rev/secΔt = 32 s[tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]
Solving for α in (9), we get:[tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]
Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:[tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]
In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:[tex]\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)[/tex]Define the average acceleration of a particle
between two given instants.
ocean currents are always cold true or false
Fill in the graph for 50 points
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Explanation:
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.2 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.3 mm and that is perpendicular to the direction of the field.
poste en français s’il vous plaît
effect of high pitch on humans
Answer:
High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.
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Betty hits a baseball.
At which point along the baseball's path does it have the most gravitational potential energy?
Answer:
point 2
Explanation:
What are the two main processes carried out by the excretory system?
Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2
How do you find the period of a sound wave?
Answer:
Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.
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I believe you are incorrect. A weathered mountain would appear more jagged.
I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.
If this is incorrect, please, don't refrain to tell me.
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.
Answer:
[tex]1.714\ \text{A}[/tex]
Explanation:
F = Magnetic force = 0.018 N
B = Magnetic field = 0.03 T
L = Length of wire = 35 cm
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]90^{\circ}[/tex]
Magnetic force is given by
[tex]F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}[/tex]
The magnitude of the current is [tex]1.714\ \text{A}[/tex].
7. Copper can be coated on the surface of iron but not on silver? why
Answer:
Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.
Explanation:
Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose
When a wave enters a new medium from an angle, both the speed and the ________ change
a
The frequency
b
The amplitude
c
The energy
d
The angle
Answer:
B: Amplitude
Explanation:
When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.
The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.
Answer:
The angle
Explanation:
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Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
Which term does this explain?
This is a non-mathematical explanation of how nature works. It must be
supported by a large body of evidence.
Fact
Law
Theory
Hypothesis
Energy is transferred between the ocean and the air to make sure that the temperature in the air is higher than the temperature on the surface False True
3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is
swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?
Answer:
the velocity of the big fish after the launch is 6 m/s.
Explanation:
Given;
mass of the big fish, m₁ = 5 kg
velocity of the big fish, u₁ = 8 m/s
mass of the small fish, m₂ = 1 kg
velocity of the small fish, u₂ = -4 m/s
Let the final velocity of the big fish after launch = v
Apply the principle of conservation linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
5 x 8 + 1 x (-4) = v(5 + 1)
40 - 4 = 6v
36 = 6v
v = 36/6
v = 6 m/s.
Therefore, the velocity of the big fish after the launch is 6 m/s.
g Suppose you have this brilliant idea: A Ferris wheel has radial metallic spokes between the hub and the circular rim (of radius roughly 16 m). These spokes move in the magnetic field of the Earth (5e-05 T), so each spoke acts like a rotating bar in a magnetic field. The magnetic field points perpendicular to the plane of the Ferris wheel. You plan to use the emf generated by the rotation of the Ferris wheel to power the light-bulbs on the wheel. Suppose the period of rotation for the Ferris wheel is 90 seconds. What is the magnitude of the induced emf between the hub and the rim
Answer:
[tex]4.46\times 10^{-4}\ \text{V}[/tex]
Explanation:
B = Magnetic field = [tex]5\times 10^{-5}\ \text{T}[/tex]
r = Radius of rim = 16 m
t = Time = 90 seconds
A = Area of rim = [tex]\pi r^2[/tex]
EMF is given by
[tex]\varepsilon=\dfrac{BA}{t}\\\Rightarrow \varepsilon=\dfrac{5\times 10^{-5}\times \pi\times 16^2}{90}\\\Rightarrow \varepsilon=0.000446=4.46\times 10^{-4}\ \text{V}[/tex]
The magnitude of the induced emf between the hub and the rim is [tex]4.46\times 10^{-4}\ \text{V}[/tex].
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
Consider a long, thin rod with a length of 3 m rotating about it's end. This rod has a moment of inertia of 12 kg·m2 about this pivot.
What is the mass of the rod? Give your answer in kilograms to two decimal places.
Answer:
The mass of the rod is 16 kg.
Explanation:
Given that,
The length of a rod, L = 3 m
The moment of inertia of the rod, I = 12 kg-m²
We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :
[tex]I=\dfrac{ML^2}{12}[/tex]
Where
M is mass of the rod
[tex]M=\dfrac{12I}{L^2}\\\\M=\dfrac{12\times 12}{(3)^2}\\\\M=16\ kg[/tex]
So, the mass of the rod is 16 kg.