Answer:
4N
Explanation:
a = (7-3)/5 = 0.8m/s^2
F = ma = (5)(0.8) = 4 Newtons
For the circuit in the previous part, the current flowing in the wire between the positive terminal of the battery and the resistor is ___________ the current flowing between the resistor and the negative terminal of the battery. For the circuit in the previous part, the current flowing in the wire between the positive terminal of the battery and the resistor is ___________ the current flowing between the resistor and the negative terminal of the battery. less than equal to greater than
Answer:
Explanation:
because it is godhood
A 45.0 kilogram boy is riding a 15.0-kilogram bicycle with a speed of 8.00 meters per second. What is the combined kinetic energy of the boy and the bicycle? A)480.J B)240.0J C)1920J D)1440J
Answer:
1920Joules
Explanation:
The formula for calculating the kinetic energy of a body is expressed as;
KE = 1/2 mv²
m isthe mass
V is the speed
For the two masses, the combined KE is expressed as;
KE = 1/2(m1+m2)v²
KE = 1/2(45+15)(8)²
KE = 1/2 * 60 * 64
KE = 30 * 64
KE = 1920J
Hence the combined kinetic energy of the boy and the bicycle is 1920Joules
The combined kinetic energy of the boy and the bicycle is of 1920 J.
Given data:
The mass of boy is, m = 45.0 kg.
The mass of bicycle is, M = 15.0 kg.
The speed of bicycle is, v = 8.00 m/s.
The kinetic energy of an object is defined as the energy possessed by an object by virtue of motion of object. The combined kinetic energy of the boy-bicycle system is given as,
[tex]KE = \dfrac{1}{2}(m+M)v^{2}[/tex]
Solve by substituting the values as,
[tex]KE = \dfrac{1}{2}(45+15) \times 8^{2}\\\\KE = 1920 \;\rm J[/tex]
Thus, we can conclude that the combined kinetic energy of the boy and the bicycle is of 1920 J.
Learn more about the concept of kinetic energy here:
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You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
(a) The speed of car A just before the collision is 51.58 mph.
(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.
What is collision?
The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.
Given data -
The mass of car A is, mA = 1500 kg.
The mass of car B is, mB = 2100 kg.
The length of the skid mark is, d = 7.30 m.
The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].
(a)
The combined kinetic energy of both cars is,
[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]
Applying the work-energy principle as,
Work done due to kinetic friction = Combined kinetic energy of cars
[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]
Converting into mph as,
[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]
To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 21.49
v₁ = 51.58 mph
Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.
(b)
With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,
= 51.58 - 35 = 28.63 mph.
= 16.58 mph
Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.
Learn more about the average speed here:
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Wind is formed when hot air rises and cool air sinks. This process is called __________
conduction
insulation
radiation
convection
plsssssssss answer this correct i put in 12 points and i will give you brainliest if you answer correct
Answer:
conduction is the correct answer
Which type of biological molecule would control cell activities?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
Answer:
C. Nucleic Acids.
Explanation:
Nucleic acids are thus uniquely capable of directing their own self-replication, allowing them to function as the fundamental informational molecules of the cell. The information carried by DNA and RNA directs the synthesis of specific proteins, which control most cellular activities.
Did some simple research, if that's fine and I hope this is correct as it says on the sites. Have a good one :)
What is a overly-simplified definition of Einstein's theory of general relativity?
Answer:
the laws of physics are the same for all non-accelerating observers
Explanation:
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on 92?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
-5.00 x 10-6
-5.00 x 10-6 C
91
92
93
0.500 m
0.250 m
q1 = -5.00 x 10-6 C
q2 = -5.00 x 10-6 C
q3 = -5.00 x 10-6 C
E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left
E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right
E net = 720000 - 180000 = 540000 N/C to the right
F = qE
F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N
The force on q2 is 2.7 N to the left.
The net electrostatic force on the q2 is 2.7N owards left
The equation for electrostatic force is
[tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]
where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.
the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1= -5.0 ×[tex]10^{-6}[/tex] C and q3= -5.0 ×[tex]10^{-6}[/tex] C
distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m
distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m
force due to charge q1
[tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N = 0.9N towards right
[tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left
hence net force F = [tex]F_{1}+F_{2}[/tex]
= 0.9N - 3.6N = -2.7N
F = 2.7 N towards left
Learn more about electrostatic force:
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An object swings in a horizontal circle, supported by a 1.8-m string. It swings at a speed of 3 m/s. What is the mass of the object given that the tension in the string is 90 N?
Answer:
Mass = 18 kg
Explanation:
Formula for force in centripetal motion is;
F = mv²/r
We have;
Mass; m.
Speed; v = 3 m/s
radius; r = 1.8 m
Force; F = 90 N
Thus;
Making m the subject;
m = Fr/v²
m = 90 × 1.8/3²
m = 18 kg
2) How much work is required to pull a sled 15
meters if you use 30N of force?
2 people
Explanation:
5
Select the correct answer.
What is the current in a parallel circuit which has two resistors (17.2 ohms and
22.4 ohms) and a power source of 6.0 volts?
ОА.
0.30 amps
OB.
9.8 amps
OC.
0.61 amps
D.
1.2 amps
Reset
Next
Answer:
Current in a parallel circuit = 0.61 amps (Approx)
Explanation:
Given:
Voltage V = 6 volt
Two resistors = 17.2 , 22.4 in parallel circuit
Find:
Current in a parallel circuit
Computation:
1/R = 1/r1 + 1 / r2
1/R = 1/17.2 + 1 / 22.4
R = 9.73 ohms (Approx)
Current in a parallel circuit = V / R
Current in a parallel circuit = 6 / 9.73
Current in a parallel circuit = 0.61 amps (Approx)
Four bicyclists travel different distances and times along a straight path. Which cyclist traveled with the greatest average
speed?
A
B
Cyclist 2 travels
87 min 22 s
Cyclist 4 travels
108 min 24 s
D
Cyclist 1 travels
95 m in 27 s
Cyclist 3 travels
106 m in 26 s
Answer:
The cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
Explanation:
Given;
Cyclist 1 travels 9 m in 27 s
Cyclist 2 travels 87 m in 22 s
Cyclist 3 travels 106 m in 26 s
Cyclist 4 travels 108 m in 24 s
Determine the average speed of the cyclists as follows;
Average speed of Cyclist 1: v = 9/27 = 0.33 m/s
Average speed of Cyclist 2: v = 87/22 = 3.96 m/s
Average speed of Cyclist 3: v = 106/26 = 4.08 m/s
Average speed of Cyclist 4: v = 108/24 = 4.5 m/s
Therefore, the cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the angular velocity of the toy?
Explanation:
angular velocity = velocity/radius
= 5/2
= 2.5 rad/s
We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an external force (or non-conservative force), there will be a change in the total mechanical energy of the object. If only internal forces are doing work then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. Think of a real-life situation where we make use of this conservation of mechanical energy (where we can neglect external forces for the most part). Describe your example and speak to both the kinetic and potential energy of the motion.
Answer:
* roller skates and ice skates.
* roller coaster
Explanation:
One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.
Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.
The only way that heat can travel through outer space is ______
convection
radiation
conduction
none of the above
plssssssssssss answer correctly
Which of these would have the highest temperature?
ice
· Water
water vapor
Answer:
water vapor
Explanation:
did assignment on edge
In the legend of William Tell, Tell is forced to shoot an apple from his son's head for failing to show respect to a high official. In our case, let's say Tell stands 8.7 meters from his son while shooting. The speed of the 144-g arrow just before it strikes the apple is 20.4 m/s, and at the time of impact it is traveling horizontally. If the arrow sticks in the apple and the arrow/apple combination strikes the ground 8 m behind the son's feet, how massive was the apple
Answer:
M = 0.31 kg
Explanation:
This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved
let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple
initial instant. Just before the crash
p₀ = m v₁ + M 0
instant fianl. Right after the crash
p_f = (m + M) v
p₀ = p_f
m v₁ = (m + M) v
v =[tex]\frac{m}{m+M} \ v_1[/tex] (1)
now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set
x = v t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground, its height is y = 0 and as it comes out horizontally, [tex]v_{oy} = 0[/tex]
0 = h - ½ g t²
t² = 2h / g
For the solution of the exercise, the height of the child must be known, suppose that h = 1 m
t = [tex]\sqrt{ \frac{ 2 \ 1}{9.8} }[/tex]
t = 0.452 s
let's find the initial velocity
v = v / t
v = 8 / 0.452
v = 17.7 m / s
From equation 1
v = m / (m + M) v₁
m + M = [tex]m \ \frac{v_1}{v}[/tex]
M = m + m \ \frac{v_1}{v}
we calculate
M = 0.144 + 0.144 [tex]\frac{20.4}{17.7}[/tex]
M = 0.31 kg
What happens to the energy of a rubber band when it is stretched?
A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 75.0 kg construction worker stands at the far end of the beam. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What is the magnitude of the torque about the point where the beam is bolted into place?
Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
how does temperature affect brownian motion
Answer: This (random) thermal motion of the particles due to the temperature is also called Brownian motion. ... The higher the temperature, the faster the diffusion will be, because the stronger the molecule movement and thus the “mixing”.
Explanation:
What happens to solar radiation when it is absorbed
Answer:
Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere
1. An object with a mass of 5 kg is pushed by a force of 10 N. What is the object's acceleration?
Answer:2m/s^2
Explanation:
a=f/m
If the net force acting on an object is 0 N, you can be sure that the forces acting on the object are
A. balanced B.Unbalanced C. acting at the same direction
I think the answer would be A.
After all, it is 0 which is technically a dead center number meaning that the net should be balanced and still.
Hope this helps and have a nice day.
-R3TR0 Z3R0
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of A is 8.00 times the mass of B, and the energy stored in the spring was 73 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. (a) Once that transfer is complete, what is the kinetic energy of particle A
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
a toy train is moved towards a magnet that cannot move. what happens to the potential energy in the system of magnets during the movement
Answer:
Dakota moves a magnetic toy train toward a magnet that cannot move. What happens to the potential energy in the system of magnets during the movement? The potential energy increases because the train moves against the magnetic force. The potential energy decreases because the train moves against the magnetic force.
I hope this helps you :)
An elevator suspended by a vertical cable is moving downward at a constant speed. The tension in the cable must be A) greater than the weight of the elevator. B) less than the weight of the elevator. C) equal to the weight of the elevator.
Answer:
(C) because the elevator is not accelerating
Note F = M a = M g (the resultant force on the elevator is due to gravity)
or Fup = Fc the force exerted on the elevator by the cable
and Fdown = Fe the force exerted on the elevator by gravity
F = M a = Fup - Fdown = zero resultant force on elevator
Are we within earths Roche limit
Answer: Closer to the Roche limit, the body is deformed by tidal forces. Within the Roche limit, the mass' own gravity can no longer withstand the tidal forces, and the body disintegrates. hope this helps can u give me brainliest
Explanation:
A student picks up two spheres that are the same size. One is made of steel and the other is made of aluminum. The student notices that the steel sphere feels much heavier than the aluminum sphere. He then holds one sphere in each hand at eye level and lets go of them at the same time. They fall to the floor. Which ball, if any, will hit the ground first and for what reason
Answer:
They will fall at the same time. This is because gravity accelerates all objects at the same speed, Earth's gravity being approximately 9.8m/s²
They'll both fall at the same time. This is because gravity accelerates everything at the same rate, with Earth's gravity being approximately 9.8 m/s2.
What is gravity with some instances?The energy that holds the gases inside the sun together. the force that causes a ball to fall after being thrown into the air the force that causes a car to coast downhill even when the gas pedal is not depressed the force that causes a glass to shatterGravity, also known as gravitation, affects all material objects in the universe. Gravity attracts any two objects or particles with nonzero mass toward one another. Gravity affects everything from subatomic particles to galaxy clusters. Gravity is the attraction force between two objects. It's what causes things to fall and keeps us from floating away into space. Gravity is a fundamental natural force.To learn more about gravity, refer to:
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Se desea elevar un cuerpo de 1000 kg utilizando una elevadora hidráulica de plato grande circular de 50 cm de radio y plato pequeño circular de 8cm de radio, calcula: a) El peso del cuerpo. b) Cuanta fuerza hay que hacer en el émbolo pequeño, c) Si el émbolo pequeño desciende 60 cm, ¿qué volumen de fluido desplaza hacia el émbolo mayor?
Answer:
a) [tex]W=9810\: N[/tex]
b) [tex]F_{1}=251.14\: N[/tex]
c) [tex]V_{g}=0.012\: m^{3}[/tex]
Explanation:
a)
El peso del cuerpo es:
[tex]W=mg[/tex]
g es la gravedad (9.81 m/s²)
[tex]W=1000*9.81[/tex]
[tex]W=9810\: N[/tex]
b)
Usando el principio de Pascal tenemos:
[tex]P_{1}=P_{2}[/tex]
y la presion es la fuerza sobre el area.
[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}[/tex]
F(1) es la fuerza aplicada en el embolo pequeñoA(1) es el area del disco pequeñoF(2) es la fuerza aplicada en el embolo grandeA(2) es el area del disco grandeDespejando F(1):
[tex]F_{1}=F_{2}\frac{A_{1}}{A_{2}}[/tex]
el area del plato es: [tex]A=\pi R^{2}[/tex]
[tex]F_{1}=F_{2}\frac{\pi R_{1}^{2}}{\pi R_{2}^{2}}[/tex]
[tex]F_{1}=F_{2}\frac{R_{1}^{2}}{R_{2}^{2}}[/tex]
F(2) es el peso del cuerpo de 1000 kg (W)
[tex]F_{1}=9810\frac{8^{2}}{50^{2}}[/tex]
Por lo tanto, la fuerza que se debe hacer es:
[tex]F_{1}=251.14\: N[/tex]
c)
Como tenemos un sistema cerrado el volumen de agua que desciende por el embolo pequeño debe ser igual al que sube por el grande, por lo tanto:
[tex]V_{p}=V_{g}[/tex]
Vp es el volumen de agua en el émbolo pequeño
Vg es el volumen de agua en el émbolo grande
Como sabemos que son cilindros (V=πR²h)
[tex]\pi R^{2}h=V_{g}[/tex]
Entonces el volumen del émbolo mayor será:
[tex]V_{g}=\pi 0.08^{2}0.6[/tex]
[tex]V_{g}=0.012\: m^{3}[/tex]
Espero te haya sido de ayuda!
In 2014, physicists from FOM Foundation at the University of Amsterdam introduced a new hypothesis of how the Pyramids at Giza were built. The group of physicists suggestedthat ancient Egyptians wetted sand in an effort toreduce friction and then pulled the 3000 kg stoneblocks to their final resting place. 15 men couldmove a block at a rate of 0.5m/sby pulling a largerope angled at 30owith respect to the plane anda tension of 7,200 N.
Required:
a. What is net work done on block?
b. What is speed of blck after it moved .25m?
c. What is work done by block if kinetic friction coefficient is 0.3?
d. What is net work including friction?
Answer:
The correct answer is:
(a) 0
(b) 0.5 m/s
(c) 7740 N
(d) 0
Explanation:
The given values are:
mass,
m = 3000 kg
Tension,
T = 7,200 N
Angle,
= 30°
(a)
Even as the block speed becomes unchanged, the kinetic energy (KE) will adjust as well:
⇒ [tex]\Delta K =0[/tex]
By using the theorem of energy, the net work done will be:
⇒ [tex]\Delta K =0[/tex]
(b)
According to the question, After 0.25 m the block is moving with the constant speed
= 0.5 m/s.
(c)
The given kinetic friction coefficient is:
u = 0.3
The friction force will be:
= [tex]u(mg-Tsin30^{\circ})[/tex]
On substituting the values, we get,
= [tex]0.3[(3000\times 9.8)-(7200\times 0.5)][/tex]
= [tex]0.3[29400-3600][/tex]
= [tex]0.3\times 25800[/tex]
= [tex]7,740 \ N[/tex]
(d)
On including the friction,
The net work will be:
⇒ [tex]\Delta K=0[/tex]
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.685 T, the strip was 0.107 mm thick, the current along the strip was 2.25 A, and the Hall voltage between the strip's edges was 2.59 mV.Find the density nof mobile electrons in the material. The elementary charge is 1.602×10−19 C.
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³