(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that
∑ F = s - mg = 0 ==> s = mg = 617.4 N
If the spring is compressed 0.015 m, then the spring constant k is such that
617.4 N = k (0.015 m) ==> k = 41,160 N/m ≈ 41 kN/m
(2) The total mass of the passengers is
24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg
so that if everyone is at rest, the spring is compressed a distance x such that
kx = (1717 kg) g ==> x ≈ 0.41 m
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?
Answer:
28 j
Explanation:
because when you add you get 28
Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4 mm. What is the magnitude of the electrostatic force experienced by charge 2
Answer: [tex]247.12\ N[/tex]
Explanation:
Given
Magnitude of the charges
[tex]q_1=4.47\times 10^{-6}\ C[/tex]
[tex]q_2=1.86\times 10^{-6}\ C[/tex]
Distance between them [tex]d=17.4\ mm[/tex]
As both charges are of same sign, they must repel each other
Force experienced by second charge is
[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]
Thus, charge 2 experience a force of [tex]247.12\ N[/tex]
Answer:
The force between the two charges is 247.15 N.
Explanation:
Charge, q = 4.47 x 10^-6 C
charge, q' = 1.86 x 10^-6 C
distance, d = 17.4 mm
Let the force is F.
The force is given by the Coulomb's law:
[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
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What is the name of the invisible line that runs
down the center of the axial region?
Answer:
An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.
Explanation:
The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.
A stone is dropped from a bridge. It takes 4s to reach the water below. How high is the bridge above the water?
Answer:
height is 78.4m
Explanation:
h=u.t + 0.5.g.t^2
= 0 + 0.5x9.8x4^2
= 78.4m
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you
Answer:
44 N
Explanation:
Given that,
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.
It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.
Why must scientists be careful when studying
nanotechnology?
Answer:
When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.
Hope it helps u:)
The position of the image obtained by convex lens when object is kept beyond 2F1(F: principal focus of the convex lens)
A. between F2 and 2F2
B. at 2F2
C. beyond 2F2
D. at infinity
Answer:
Between F2 and 2F2
Explanation:
Diagram attached from Teachoo.
Link to website if you need to refer
https://www.teachoo.com/10838/3118/Convex-Lens---Ray-diagram/category/Concepts/
A 0.4 m long solenoid has a total of 356 turns of wire and carries a current of 79 A. What is the magnitude of the magnetic field at the center of the solenoid?
Answer:
B = 0.088 T
Explanation:
Given that,
The length of a solenoid, l = 0.4 m
No. of turns of wire, N = 356
Current, I = 79 A
We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.
[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]
So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.
Một ống dây điện thẳng dài có lõi sắt, tiết diện ngang của ống S = 20 cm2
, chiều dài
1 m, hệ số tự cảm L = 0,44 H. Cường độ từ trường trong ống dây là H = 0,8.103 A/m. Từ
thông gửi qua tiết diện ngang của ống bằng
3
0
1,6.10 Wb
. Cường độ dòng điện chạy
qua ống dây là
Answer:
sgsbssbduebubbeeifirjeirneejrbb8m!keoejr
d
iejejjeiie
1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Answer:
t = 1.27 x 10⁹ s
Explanation:
First, we will find the volume of the wire:
Volume = V = AL
where,
A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²
L = Length of wire = 150 km = 150000 m
Therefore,
V = 47.12 m³
Now, we will find the number of electrons in the wire:
No. of electrons = n = (Electrons per unit Volume)(V)
n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)
n = 3.97 x 10³⁰ electrons
Now, we will use the formula of current to find out the time taken by each electron to cross the wire:
[tex]I =\frac{q}{t}[/tex]
where,
t = time = ?
I = current = 500 A
q = total charge = (n)(chareg on one electron)
q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)
q = 6.36 x 10¹¹ C
[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]
Therefore,
t = 1.27 x 10⁹ s
Electric field is always perpendicular to the equipotential surface.
a. True
b. False
Answer:
a: true.
Explanation:
We can define an equipotential surface as a surface where the potential at any point of the surface is constant.
For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.
Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.
Now we want to see if the electric field is always perpendicular to these equipotential surfaces.
You can see that in the two previous examples this is true, but let's see for a general case.
Now suppose that you have a given field, and you have a test charge in one equipotential surface.
So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.
But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.
Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.
The correct option is a.
How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A) 0.50c
B) 0.65c
C) 0.78c
D) 0.87c
Answer:
Option (D) is correct.
Explanation:
Let the speed is v.
[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]
Option (D) is correct.
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
A 1,071.628 N painter needs to climb d=1.926 m up a ladder (measured along its length from the point where the ladder contacting the ground), without the ladder slipping. The uniform ladder is 12.014 m long and weighs 250 N. It rests with one end on the ground and the other end against a perfectly smooth vertical wall. The ladder rises at an angle of theta=51.96 degrees above the horizontal floor. What is friction force in unit of N that the floor must exert on the ladder? Use g = 10 m/s2 if you need to .
The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N
The known values are;
The weight of the painter = 1,071.628 N
The height to which the painter needs to climb along the ladder = 1.926 m
The length of the ladder = 12.014 m
The weight of the ladder = 250 N
The points where one of the ladder's ends is resting = On the ground
The points where the other end of the ladder is resting = A perfectly smooth wall
The angle with which the ladder rises above the horizontal floor = 51.96°
The acceleration due to gravity, g ≈ 10 m/s²
The unknown values include;
The friction force that the floor must exert on the ladder
The strategy to be used;
At equilibrium, the sum of moments about a point is zero
Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;
At equilibrium, the sum of clockwise, [tex]M_{CW}[/tex], moment about P = The sum of the counterclockwise, [tex]M_{CCW}[/tex]moment about P
[tex]\mathbf{M_{CCW}}[/tex] = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
[tex]\mathbf{M_{CW}}[/tex] = 12.014 × cos(51.96°) × [tex]\mathbf{F_N}[/tex]
Where;
[tex]\mathbf{F_N}[/tex] = The normal reaction of the of the ground on the end of the ladder that rests on the floor
[tex]\mathbf{M_{CCW}}[/tex] = [tex]\mathbf{M_{CW}}[/tex]
∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × [tex]F_N[/tex]
We get;
6,665.3068846 N·m = 7.40316448688 m × [tex]F_N[/tex]
[tex]\mathbf{F_N}[/tex] = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N
The normal reaction of the floor on the ladder, [tex]\mathbf{F_N}[/tex] = 900.332135 N
Taking moment about the point the ladder rests on the floor, R, gives;
[tex]M_{CCW}[/tex] = 12.014 × sin(51.96°) × [tex]F_W[/tex]
Where;
[tex]\mathbf{F_W}[/tex] = The normal reaction at the wall
[tex]M_{CW}[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
At equilibrium, we have, [tex]M_{CCW}[/tex] = [tex]M_{CW}[/tex]
Therefore;
12.014 × sin(51.96°) × [tex]F_W[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
9.46199511627 m × [tex]F_W[/tex] = 2,197.22861125 N·m
[tex]F_W[/tex] = 2,197.22861125 N·m/(9.46199511627 m)
The reaction of the wall, [tex]\mathbf{F_W}[/tex] = 232.216206 N
We note that also at equilibrium, the sum horizontal forces = 0
The horizontal forces acting on the ladder = The normal reaction on the, [tex]F_W[/tex] wall and the friction force on the ground, [tex]\mathbf{F_f}[/tex]
∴ At equilibrium; [tex]\mathbf{F_W}[/tex] + [tex]\mathbf{F_f}[/tex] = 0
[tex]\mathbf{F_f}[/tex] = -[tex]\mathbf{F_W}[/tex]
[tex]\mathbf{F_W}[/tex] = 232.216206 N
Therefore;
The frictional force in unit of N that the floor must exert on the ladder, [tex]\mathbf{F_f}[/tex] = 232.216206 N ≈ 232.216 N.
(The coefficient of friction, μ = [tex]\mathbf{F_N}[/tex]/[tex]\mathbf{F_W}[/tex] = 900.332135/232.216206 ≈ 3.877).
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3. The figure below shows the motion of a car. It starts from the origin, O travels 8m
towards the east and then 12m towards the west.
D
8m.
X
X-8
12m.w
()What is the net displacement D from the origin to the final position?
(ii) What is the total distance travelled by the car?
Answer:
i. -4m
ii. 20m
Explanation:
The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)
Total distance = 8m going east + 8m back to origin + 4m west = 20m
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
find the exit angle relative to the horizontal in an isosceles triangle with 36 °
what
what
what
what
sorry
sorrry
sorry
At what angle torque is half of the max
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted
between S and and q is __ the force exerted between S and p.
1/6
6 times
1/36
36 times
Answer:
Fq = k q Q / R^2
Fp = k q Q / (6 R)^2
The force exerted between S and p is 1 / 36 of that between S and q
or the force between S and q is 36 times that between S and p.
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
When placed 1.18 m apart, the force each exerts on the other is 11.2 N and is repulsive. What is the charge on each
Answer:
[tex]q=41.62\ \mu C[/tex]
Explanation:
Given that,
Force between two objects, F = 11.2 N
Distance between objects, d = 1.18 m
We need to find the charge on each objects. The force between charges is as follows :
[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]
So, the charge on each sphere is [tex]41.62\ \mu C[/tex].
Your forehead can withstand a force of about 6.0 kN before it fractures. Your cheekbone on the other hand can only handle about 1.3 kN before fracturing. If a 140 g baseball hits your head at 30.0 m/s and stops in 0.00150 s,
Required:
a. What is the magnitude of the ball's acceleration?
b. What is the magnitude of the force that stops the baseball?
c. What force does the baseball apply to your head? Explain?
d. Are you in danger of a fracture if the ball hits you in the forehead?
Answer:
Explanation:
a)
Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s
v = u + a t
0 = 30 m/s + a x .0015 s
a = - 30 / .0015
= - 20000 m / s²
b )
Magnitude of force = m x a
= .140 kg x 20,000 m / s²
= 2800 N = 2.8 kN.
c )
The force applied by baseball = 2.8 kN .
d )
Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.
What is inertia of motion?
Explanation:
Inertia of motion
It is also known as Newton's first law of motion.
It states that,
An object remains in a state of rest or of uniform motion in a straight line unless compelled to change its state by an applied external force.
3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following correctly describes the tension in the three rope segments?
a. T1< T2 < T3
b. T1< T2 = T3
c. T1 = T2 = T3
d. T1> T2 > T3
please help.show how and which?
see attachment for more detail.
Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.
Let's evaluate each tension.
Case T₃.
[tex] T_{3} - W_{3} = 0 [/tex]
For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.
[tex] T_{3} = W_{3} [/tex]
Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:
[tex] T_{3} = W_{3} = mg [/tex] (1) Case T₂.
[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]
[tex] T_{2} = T_{3} + W_{2} [/tex] (2)
By entering W₂ = 2mg and equation (1) into eq (2) we have:
[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]
Case T₁.
[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]
[tex] T_{1} = T_{2} + W_{1} [/tex] (3)
Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:
[tex] T_{1} = 3mg + 3mg = 6mg [/tex]
Therefore, the correct option is d: T₁ > T₂ > T₃.
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Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]
First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:
Mass m:
[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)
Mass 2m:
[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)
Mass 3m:
[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)
The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown
Answer: [tex]283.2\times 10^{-9}\ nC[/tex]
Explanation:
Given
Cross-sectional area [tex]A=0.4\ cm^2[/tex]
Dielectric constant [tex]k=4[/tex]
Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]
Distance between capacitors [tex]d=5\ mm[/tex]
Maximum charge that can be stored before dielectric breakdown is given by
[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]
Answer:
The maximum charge is 7.08 x 10^-8 C.
Explanation:
Area, A = 0.4 cm^2
K = 4
Electric field, E = 2 x 10^8 V/m
separation, d = 5 mm = 0.005 m
Let the capacitance is C and the charge is q.
[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]
(a) State Hook's law. [2]
(b) The walls of the tyres on a car are made of a rubber
compound. The variation with stress of the strain of a
specimen of this rubber compound is shown in Fig. 1.2.
As the car moves, the walls of the tyres end and straighten
continuously. Use Fig. 1.2 to explain why the walls of the
tyres become warm
[3]
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Fig. 1.2.
Hooke's law gives the relationship between the force applied and observed compression and expansion
(a) Hooke's law states that force applied is proportional to the deformation of an object
(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre
The reason for the above explanation are as follows;
(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F
Mathematically, we get;
F = k × ΔL
[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]
Given that the material cross sectional area = A, and the original length of the material = L, we get;
F/A = Stress = σ, ΔL/L = strain = ε
[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]
Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material
(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation
Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve
As the tyre straightens, the path of the stress change curve is given by the lower curve
The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening
Energy absorbed < Energy released
The balance energy is transformed into other forms of energy, including heat energy, which is observed by the raising in temperature, warming, of the tyre
Energy absorbed = Energy released + Heat energy
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