Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = [tex]\frac{2}{5}[/tex] m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²
m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
[tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]
Tasks
Task 2 - Compare and contrast the use of D'Alembert's principle with
the principle of conservation of energy to solve an
engineering problem
A motor vehicle having a mass of 800 kg is at rest on an incline of 1 in 8 when the
brakes are released. The vehicle travels 30 m down the incline against a constant
frictional resistance to motion of 100 N where it reaches the bottom of the slope.
a) Using the principle of conservation of energy, calculate the velocity of the
vehicle at the bottom of the incline.
b) Using an alternative method that does not involve a consideration of energy,
cacluate the velocity of the vehicle at the bottom of the incline.
c) Discuss the merits of the two methods you have used for parts a) and b) of
this question. Justify the use of an energy method for these types of
problems.
Answer:
NE DIYON INGILIZ MISIN SEN
A box of bananas weighing 51.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.46 and the coefficient of kinetic friction is 0.23. Part A If no horizontal force is applied to the box and the box is at rest, how large is the frictional force exerted on the box by the surface
static friction force = The coefficient of static friction * normal force
static friction force = 0.46 * 51 = 23.46 N
kinetic friction force = The coefficient of kinetic friction * normal force
kinetic friction force = 0.23 * 51 = 11.73 N
the applied force acting on the object must be more than 23.5 N if the object was stationary to move it and must be more than 11.7 N during the movement to keep the object moving
A pendulum has a period of 5.14s and a length of 0.25m. What is the acceleration
due to gravity? *
Answer:
Acceleration due to gravity, g = 2.68m/s²
Explanation:
Given the following data;
Period = 5.14s
Length = 0.25m
To find acceleration due to gravity, g;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]
[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]
[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]
[tex] 0.8180 = \sqrt {0.25g} [/tex]
Taking the square of both sides
[tex] 0.8180^{2} = 0.25g [/tex]
[tex] 0.6691 = 0.25*g[/tex]
[tex] g = \frac {0.6691}{0.25} [/tex]
Acceleration due to gravity, g = 2.68m/s²
A major source of water pollution comes from that washes chemicals and other pollutants from improperty managed land.
HELP ASAP PLS
A balloon with a positive charge will stick to a wall that has a negative charge.
What force causes this?
A. Gravity
B. Electric force
C. Magnetic force
D. Air gesistance
Given a volume of 1000. Cm^3 of an ideal gas at 300 k, what volume would iy occupy at a temperature of 600 k
Answer:2000 cm³
Here, pressure remains constant.
So, b the gas law
V/V' = T/ T'
1000 / V' = 300 / 600
V' = 2000 cm³
Explanation:also pls mark brainliest
Which object would have the greatest acceleration?
Answer:
D
Explanation:
A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.
Which is a valid velocity reading for an object?
45 m/s
45 m/s north
O 0 m/s south
O 0 m/s
Answer:45 m/s north
Explanation:
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
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What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C
Answer:
[tex]q=-2.26\times 10^{-5}\ C[/tex]
Explanation:
Given that,
The mass of a particle, m = 1.48 g = 0.00148 kg
The electric field, E = 640 N/C
We need to find the charge of the particle when placed in a downward-directed electric field.
The force of gravity is balanced by the electric force such that,
mg = qE
Where
q is the charge of the particle
[tex]q=\dfrac{mg}{E}\\\\q=\dfrac{0.00148\times 9.8}{640}\\\\q=2.26\times 10^{-5}\ C[/tex]
q must be negative, the force must be upward (opposite direction of the electric field).
The driver of a 3000 lb. car, coasting down a hill, sees a red light at the bottom, and must stop. His speed when he applies the brakes is 60 mph, and he is 100 feet (vertically) above the bottom of the hill. (a)How much energy as heat must be dissipated by the brakes if we neglect wind resistance and other frictional effects
Answer:
Explanation:
60 mph = 60 x 1760 x 3 / (60 x 60) ft /s
speed of car , v = 88 ft /s
kinetic energy of car = 1/2 m v²
= .5 x 3000 x 88²
= 11616 x 10³ poundal - foot
Potential energy = mgh
= 3000 x 32 x 100
= 9600 x 10³ poundal - foot
Total energy = potential energy + kinetic energy
= ( 11616 + 9600 )x 10³
= 21216 x 10³ poundal - foot .
This energy is dissipated as heat when brakes are applied on the car to stop the car .
What composes about 71% of Earth's outermost layer?
A
oceanic crust
B
asthenosphere
С
lithosphere
D
continental crust
Answer:
A. oceanic crust
Explanation:
I remember that the ocean is said to cover 71% of the Earth's surface. If you look at a globe, notice that most all the surface is blue like the ocean.
One other note: the surface is the Earth's outermost layer. Think of it this way: surface implies the top of something, something exposed to the outside.
Therefore, the answer is A. Hope this helps you understand the question more! Have a great day, 'kay?
. Why is it harder to stop an elephant accelerating at 1m/s2 than a rabbit accelerating at 1m/s2
(10 Points)
the elephant has more mass
the rabbit is faster
the rabbit has more mass
the elephant is faster
Answer:
this is about momentum p=mv
A, the elephant has more mass
Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
Answer:
the Potential Energy is 2.304 × 10⁻¹⁶ J
Explanation:
Given the data in the data in the question;
The expression for the electric potential energy between the charges can be expressed as follows;
PE = qV ------equ 1
where q is the charge and V is the electric potential
Also the formula for electric potential due to point a point in a field is;
V = kq / r -------equ 2
where k is the electrostatic constant and r is the distance form the charged particle
input equation 2 into 1
PE = q × kq / r
PE = kq²/r ------- equ 3
so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )
PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²
PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²
PE = 2.304 × 10⁻¹⁶ J
Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
how can you rewrite the force formula (f=ma) to solve the acceleration?
The force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
From Newton's 2nd law of motion, we can write that:
Force = mass × acceleration
⇒ acceleration = force/mass.
Hence, the force formula can be rewritten to solve the acceleration as:
acceleration = force/mass.
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If the coefficient of kinetic friction is 0.43 for a box sliding across your lab table and the
box weighs 7.4 N, what is the force of kinetic friction?
Answer:
3.2N
Explanation:
Given parameters:
Coefficient of kinetic friction = 0.43
Weight of box = 7.4N
Unknown:
Force of kinetic friction = ?
Solution:
The force of kinetic friction is given as:
Force of kinetic friction = UN
U is the coefficient of friction
N is the weight
Force of kinetic friction = 0.43 x 7.4 = 3.2N
Two balls are thrown against a wall with the same velocity. The first ball is made of rubber and bounces straight back with some non-zero speed. The second ball is made of clay and sticks to the wall after impact. If we assume the collision time was the same for each ball, which ball experienced a greater average acceleration during the collision with the wall? A. the average acceleration was the sameB. the clay ball C. there is not enough information D. the rubber ball
Answer:
A. the average acceleration was the same
Explanation:
Acceleration is calculated by finding the difference of the initial velocity from the final velocity (on impact, usually 0) and then dividing by the amount of time that took place. If we assume that both balls were thrown at the same initial force, and ended up hitting the wall at the same time then we can say that the average acceleration was the same. If the initial velocity was not the same then we would need the initial velocity of each ball in order to calculate the acceleration of each object and determine which had a greater acceleration.
A car has a mass of 2000 kg and accelerates at 2 meters per second per second. What is the magnitude of the net force exerted on the car?
Hello!
[tex]\large\boxed{4000 N}[/tex]
Use the following equation to solve for the net force (N):
∑F = m × a
Plug in the given mass (kg) and speed (m)
∑F = 2000 * 2
Simplify:
∑F = 4000 N
Climbing a tree would
A. decrease your potential energy
B. increase your kinetic energy
C. increase your potential energy
Answer:
C Increase your potential energy
Explanation:
Because if you start falling your potential energy would convert to kinetic energy. So you would get potential energy climbing up a tree
Answer:
I think it's potential energy
Substance X is in a chamber (Chamber 1) containing 10L of fluid. The concentration of substance X in this chamber is 100mmol/L. A solid division separates this chamber from another chamber (Chamber 2) containing 5L of fluid. You remove this division leaving only a membrane separating the two chambers. This membrane is permeable to Substance X and nothing else. Which of the following statements are true?There are several answers to this question, but only one is giving me issues. One of the correct answers is: "The concentration of Substance X in Chamber 2 will never exceed 66mmol/L." I do not understand how they got this as I did a ratio and came up with 50 mmol/L instead. Could someone please point out how it is 66?
Answer:
Explanation:
concentration of substance X in chamber 1 = 100 mmol/L
total volume of chamber 1 = 10 L
total mass of substance X in chamber 1 = 100 x 10 mmol = 1000 mmol .
When the two chamber is joined , total volume of both the chamber
= 10 L + 5 L = 15 L .
In the volume of 15 litre , substance x is uniformly distributed because it is permeable .
concentration of substance X = mass of X / total volume = 1000 mmol / 15
= 66.67 mmol / L
Hence ,
"The concentration of Substance X in Chamber 2 will never exceed 66mmol/L."
An electric charge at rest produce
Answer:
Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.
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To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal length of the eyepiece lens should be ________. To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal length of the eyepiece lens should be ________. small; small small; large large; small large; large
Answer:
Large; small.
Explanation:
A telescope can be defined as an optical instrument or device which comprises of a curved mirror and lenses used for viewing distant objects i.e objects that are very far away such as stars and other planetary bodies. The first telescope was invented by Sir Isaac Newton.
To have the highest magnification in a telescope, the focal length of the objective lens should be large and the focal length of the eyepiece lens should be small.
This ultimately implies that, the eyepiece lens has a small focal length while the objective lens has a large focal length.
Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel (stirrer) that transfers energy to the air at a constant rate of 10 W for 1h. If no changes in kinetic or potential energy occur, determine a) The specific volume at the final state, in m3 /kg. b) The energy transfer by work, in kJ. c) The change in specific internal energy of the air, in kJ/kg.
Answer:
[tex]0.3\ \text{m}^3/\text{kg}[/tex]
[tex]36\ \text{kJ}[/tex]
[tex]18\ \text{kJ/kg}[/tex]
Explanation:
V = Volume of air = [tex]0.6\ \text{m}^3[/tex]
P = Power = 10 W
t = Time = 1 hour
m = Mass of air = 2 kg
Specific volume is given by
[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]
The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]
Work done is given by
[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]
The energy transfer by work, is [tex]36\ \text{kJ}[/tex]
Change in specific internal energy is given by
[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]
The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]
A heat pump is used to heat a building. The external temperature is lower than the internal temperature. The pump's coefficient of performance is 3.70, and the heat pump delivers 7.27 MJ as heat to the building each hour. If the heat pump is a Carnot engine working in reverse, at what rate must work be done to run it
Answer:
Heat pump needs 1.965 megajoules each hour to run.
Explanation:
The Coefficient of Performance ([tex]COP[/tex]), no unit, of a Carnot's heat pump is:
[tex]COP = \frac{Q_{H}}{W}[/tex] (1)
Where:
[tex]Q_{H}[/tex] - Heat received by the building, measured in megajoules.
[tex]W[/tex] - Work needed to run the heat pump, measured in megajoules.
If heat pump is a Carnot engine working in reverse, then the amount of work needed to run the heat pump is the least possible work. If we know that [tex]Q_{H} = 7.27\,MJ[/tex] and [tex]COP = 3.70[/tex], then the amount needed by the heat pump each hour is:
[tex]W = \frac{Q_{H}}{COP}[/tex]
[tex]W = \frac{7.27\,MJ}{3.70}[/tex]
[tex]W = 1.965\,MJ[/tex]
Heat pump needs 1.965 megajoules each hour to run.
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
We have that the speed must be at the speed below if they want to just hit their target
From the Question we are told that
Distance [tex]d=50m[/tex]
Height [tex]h=70m[/tex]
Constant Velocity [tex]v= 7.0 m/s[/tex]
Generally the equation for the time is mathematically given as
[tex]T=\frac{h}{v}\\\\T=\frac{70}{7}\\\\T=10s[/tex]
Therefore
The velocity required to make horizontal movement is
[tex]V=\frac{d}{T}\\\\V=\frac{50}{10}\\\\V=5m/s[/tex]
Given that
Velocity on the Vertical axis is
[tex]v_y=7m/s[/tex]
Velocity on the horizontal axis is
[tex]v_x=5m/s[/tex]
Therefore resultant speed
[tex]v_r=\sqrt{v_x^2+V_y^2}\\\\v_r=\sqrt{(5)^2+(7)^2}[/tex]
[tex]v_r=8.6023m/s[/tex]
In conclusion
[tex]v_r=8.6023m/s[/tex] must be their total speed if they want to just hit their target
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It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil would it take to make a capacitorlarge enoughto hold this amount of energy if one were to use plastic garbage bag with a 2.6 x 10-5m thickness that breaks down at 610 volts as a dielectric
Answer:
L = 1.11 x [tex]10^{6}[/tex] m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x [tex]10^{5}[/tex] m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x [tex]10^{-2}[/tex] m
Length = ?
So,
we know that,
U = 1/2 C Δ[tex]v^{2}[/tex]
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ[tex]v^{2}[/tex]
52000 J = (0.5) x (C) x ([tex]610^{2}[/tex])
C = 0.28 F
And we also know that,
C = [tex]\frac{K*E*A}{d}[/tex]
E = 8.85 x [tex]10 ^{-12}[/tex]
K = 3.7
A = 0.20 x L
d = 2.6 x [tex]10^{5}[/tex] m
Plugging in the values into the formula, we get:
0.28 = [tex]\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }[/tex]
Solving for L, we get:
L = 1.11 x [tex]10^{6}[/tex] m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Which formula is used to find an objects acceleration
Answer:
a=∆v/∆t
Explanation:
The definition of Acceleration is the change in velocity in a given time. So this means you first calculate ∆v (Change in velocity), and you calculate ∆t which is the time taken to apply that change in velocity. Then you find a= ∆v/∆t. This gives us the equation of Acceleration.
Answer:
C. a=∆v/∆tExplanation:
) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tightly wound around the solenoid at its center. The mutual inductance of the coil and solenoid is At a given instant, the current in the solenoid is and is decreasing at the rate of At the given instant, what is the induced current in the coil
Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil
Answer:
6.169 μA
Explanation:
Formula for induced EMF is given by the equation;
EMF = M(di/dt). We are given;
di/dt = 2.5 A/s
M = 19μH = 19 × 10^(-6) H
Thus;
EMF = 19 × 10^(-6) × 2.5.
EMF = 47.5 × 10^(-6) V
Formula for current is;
i = EMF/R. R is resistance given as 7.7 ohms.
Thus; i = 47.5 × 10^(-6)/7.7
i = 6.169 μA
a long solid rod 4.5 cm in radius carries a uniform volume charge density. if the electric field strength at the surface of the rod (not near either end) is 16 kn/c, what is the volume charge density
Answer:
6.29 μC/m³
Explanation:
Volume charge density is the quantity of charge per unit volume.
The direction of the electric field was not specified, therefore the volume charge density (ρ) is given by:
2πRLE = ρπR²L/ε₀
ρ = 2Eε₀ / R
Where E = electric field strength = 16 kN/C = 16 * 10³ N/C, R = radius of rod = 4.5 cm = 0.045 m, ε₀ = relative permittivity of free space = 8.85 * 10⁻¹² C² / Nm²
Therefore:
ρ = 2(16 * 10³ N/C)(8.85 * 10⁻¹² C²/Nm²) / 0.045 m = 6.29 * 10⁻⁶ C/m³
ρ = 6.29 μC/m³
