Answer:
diameter of the wire = 0.05 in =0.05 /12 =4.167 *10 ^-3 ft
area of cross section of the wire = A = 22/7 * ( d /2 ) ^2 =0.786 * ( 4.167 *10 ^-3 ) ^2 =1.365 *10 ^-5 ft2
E =...
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
2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:
a.On Earth, where the acceleration due to gravity is 9.8m/s?
b. On Mars, where the acceleration due to gravity is 3.8m/s?
Answer:
a=24.5 b=9.5
Explanation:
If you wrap 150 coils of heavy wire around a big iron nail and attach the ends of the wire to a 6.0v battery, you have a A) radio B) electromagnet C) galvanometer D) ammeter
Answer:
B
Explanation:
Because of the voltage attached to the iron nail
ocean currents are always cold true or false
Rewire each of the following using the correct prefix using 2 decimal places where applicable.
a.0.00000123N
b. 417 000 000 kg
c. 246800
d. 0,00088 mm
Answer:
a. 1.2×10^-6
b. 0.42×10^9
c. 246.8×10^3
d. 88
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].
A concave lens cannot produce a real image.
A. True
B. False
Answer:
B. False
A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point.
~Hoped this helped~
~Brainiliest?~
30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Answer:
[tex]18.18\ \text{m/s}[/tex]
[tex]6.82\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of large object = 8 kg
[tex]m_2[/tex] = Mass of smaller object = 3 kg
When large mass is moving
[tex]u_1[/tex] = 25 m/s
[tex]u_2[/tex] = 0
For completely inelastic collision we have the relation
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}[/tex]
Speed of the combined mass when the larger object is moving is [tex]18.18\ \text{m/s}[/tex]
When smaller mass is moving
[tex]u_1[/tex] = 0
[tex]u_2[/tex] = 25 m/s
[tex]v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}[/tex]
Speed of the combined mass when the smaller object is moving is [tex]6.82\ \text{m/s}[/tex]
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].
To learn more about the internal energy refer to the link;
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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:
The way you change the speed of a wave is to:
a
Change it's medium
b
Change it's energy
c
Transfer it to a new position
d
Apply a force
Answer:
transfer it to a new position
Match the descriptions with the graphs !
Answer:
Graph 1 matches with B, 2 with A, and 3 with C.
Explanation:
Graph 2 shows a car whose distance part of the graph is not going up or down, while the time going up. That means that the car is stopped. Graph 1 shows a straight line, meaning that the car is traveling at a constant speed. Graph 3 is a curved line, meaning the speed of the car is changing somehow, and since the line is becoming more horizontal, the car is getting slower.
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:
The amount of torque is 0.03 N.m.
Explanation:
To find the amount of torque we need to use the following equation:
[tex] \tau = \vec {r} \times \vec{F} = rFsin(\theta) [/tex] (1)
Where:
r: is the radius = 1x10⁻² m
F: is the force = 6 N
θ: is the angle = 30°
By entering the above values into equation (1) we have:
[tex]\tau = 1 \cdot 10^{-2} m*6 N*sin(30) = 0.03 N.m[/tex]
Therefore, the amount of torque is 0.03 N.m.
I hope it helps you!
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)]
A straight wire, labeled as Wire A, lies horizontally on a tabletop and is oriented to run north-south. A conventional current of 1.0 amperes runs in the wire directed towards the north. A second wire, labeled as Wire B, is also laid on the tabletop oriented north- south. Which of the following statements is true?
a. If Wire B carries no current in it and lies to the left (west) of wire A, then it will experience an attractive force to the right towards wire A).
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
c. If Wire B carries a northward conventional current and lies to the right (east) of wire A, then it will experience a repulsive force to the right (away from Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
Answer:
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
Explanation:
Two parallel conductors experience attractive force when the current flowing in the conductors are in the same direction.
Also two parallel conductors experience repulsive force when the current flowing in the conductors are in opposite direction.
Therefore, b and d are the correct options.
b. If Wire B carries a northward conventional current and lies to the left (west) of wire A, then it will experience an attractive force to the right (towards Wire A).
d. If Wire B carries a southward conventional current and lies to the left (west) of wire A, then it will experience a repulsive force to the left (away from Wire A).
True or false? A system must contain more than one object.
Answer:
true
Explanation:
normally -No system has ever performed well with one object.
A system must contain more than one object is a true statement.
What is system?A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.
Normally, no system has ever performed well with one object.
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A consumer uses 3098 kWh in 29 days. The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity . What is the consumer's electric bill for the 29 days?
The consumer electric bill for 29 days is AED 331.31
Given that, The utility company charges AED 0.077592 per kWh for the electricity plus AED 0.029998 per kWh for the distribution of the electricity
Since, A consumer uses 3098 kWh in 29 days.
The utility company charges for electricity is,
[tex]=0.077592*3098=240.38[/tex]
The utility company charges for distribution of the electricity is,
[tex]=0.029998*3098=92.93[/tex]
So that, The consumer electric bill is,
[tex]=240.38+92.93=333.31[/tex]
Hence, the consumer electric bill for 29 days is AED 331.31
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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.
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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.
How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then
pushing it back to its original position? How much power is used if this work is done in 20 sec?
Answer:
35/2 J/s
Explanation:
Just use the 2 formulas
Work done = Force * distance moved
Power = Work done/time
WD = 7 * 50 = 350
Power = 350 / 20
= 35/2 J/s
which two options describes behaviors of particles that are related to the chemical properties of the materials
a- forming hydrogen bonds between them
b- reacting quickly with water
c- having a high mass
d- forming bonds with other atoms
Answer:
The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.
Explanation:
I took the quiz on a.pex and these were correct.
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].
Help me, 100 points to answer right, answer without context will be reported
1. In the situation below, a tractor pulls a 850 sledge along a ramp of height ℎ = 1 and large = 30 °. If the tractor applies a constant force to the sledge = 6750 , at an angle = 36.9 °, determine the total work performed by all forces on the sledge to move it along the ramp. The coefficient of kinetic friction between the sledge and the plane is = 0.3. Tip: for the calculation, remember that only the components of the windows that are parallel to the direction of travel contribute to the work. Disregard the dimensions of the sled.
2) When firing a 2 projectile at a 1.4 bloco block, initially at rest, it is observed that the projectile is stuck in the block and the system moves together for a distance = 0.1 before stop. If the coefficient of kinetic friction between the block and the surface is = 0.25, determine what was the velocity of the projectile in the instant before impact. Tip: here you must use the conservation of linear momentum and also energy, considering the work done by the frictional force
ASAP 20 POINTS!!
The air also contained a small amount of argon
As the temperature of the air decreased from 20C to -190 C the argon changed
Explain the changes in arrangement and movement of the particles of the argon as the temperature of the air decreased
Answer:
See explanation
Explanation:
Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.
As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.
Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.
a forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite direction determine the magnitude and direction of the resultant force acting on the crate
Answer:
130n on the 2nd horizontal
Explanation:
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:
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
Betty hits a baseball.
At which point along the baseball's path does it have the most gravitational potential energy?
Answer:
point 2
Explanation:
A car travels at a constant speed of 25 m/s. Find the power supplied by the engine if it can supply a maximum force of 18,000 N
Answer:
720
Explanation: