Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and high will the ball go?
Answer:
About 3.06 seconds
Explanation:
[tex]v_f=v_o+at[/tex]
Since at the peak of its trajectory, the ball will have no velocity, you can write the following equation:
[tex]0=30+(-9.81)t\\\\-30=-9.81t\\\\t\approx 3.06s[/tex]
Hope this helps!
An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate the current and voltage output when the detector is used in the photo-conductive and photovoltaic modes respectively. The reverse saturation current (Is) is 9 nA.
Answer:
I = 4.189 mA V = 0.338 V
Explanation:
In order to do this, we need to apply the following expression:
I = Is[exp^(qV/kT) - 1] (1)
However, as the junction of the diode is illuminated, the above expression changes to:
I = Iopt + Is[exp^(qV/kT) - 1] (2)
Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:
I ≅ Iopt
Therefore:
I ≅ I₀Aλq / hc (3)
Where:
I₀A = Area of diode (radiation)
λ: wavelength
q: electron charge (1.6x10⁻¹⁹ C)
h: Planck constant (6.62x10⁻³⁴ m² kg/s)
c: speed of light (3x10⁸ m/s)
Replacing all these values, we can get the current:
I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)
I = 4.189x10⁻³ A or 4.189 mA
Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:
I = Is[exp^(qV/kT) - 1]
k: boltzman constant (1.38x10⁻²³ J/K)
4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]
4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]
465,444.44 + 1 = exp(38.65V)
ln(465,445.44) = 38.65V
13.0508 = 38.65V
V = 0.338 V
Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?
Answer:
a. 63.89° in the north-southward manner
b. 2.2 sec
Explanation:
The goose is flying at 100 km/h
Air from east to west is 44 km/h
angle relative to the north-south direction for the bird to travel south will be
cos∅ = 44/100 = 0.44
∅ = [tex]cos^{-1}[/tex]0.44 = 63.89° in the north-southward manner
Speed south relative to the ground will be v
Tan 63.89 = v/100
2.04 = v/100
v = 2.04 x 100 = 204 km/hr
to cover a distance of 450 m from north to south at this speed time will be
t = d/v = 450/204 = 2.2 sec
The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approximately 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Answer:
The entropy change of the Universe that occurs is 19.346 J/K
Explanation:
Given;
temperature of the sun, [tex]T_s[/tex] = 5,300 K
temperature of the Earth, [tex]T_E[/tex] = 293 K
radiation energy transferred by the sun to the earth, E = 6000 J
The sun loses Q of heat and therefore decreases its entropy by the amount
[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]
The earth gains Q of heat and therefore increases its entropy by the amount
[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]
The total entropy change is:
[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]
Therefore, the entropy change of the Universe that occurs is 19.346 J/K
Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.
Answer:
The voltage is [tex]V = 418.60 \ Volts[/tex]
Explanation:
From the question we are told that
The area of the both plate is [tex]A = 7.00 *10^{-3} \ m^2[/tex]
The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]
The magnitude of the charge is [tex]q = 5.40 *10^{-8} \ C[/tex]
The capacitance of the capacitor that consist of the two plates is mathematically represented as
[tex]C = \frac{\epsilon _o A}{d}[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So
[tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]
[tex]C = 1.29 *10^{-10} \ F[/tex]
The potential difference between the plate is mathematically represented as
[tex]V = \frac{ Q}{C }[/tex]
[tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]
[tex]V = 418.60 \ Volts[/tex]
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.
Answer:
C) less than 129 lb.
Explanation:
Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .
If R be the reaction force
For the elevator is going downwards with acceleration a
mg - R = ma
R = mg - ma
R measures its apparent weight . Spring scale will measure his apparent weight.
So its apparent weight is less than 129 lb .
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) . Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Complete Question
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .
Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Part B
If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
Answer:
Part A
[tex]R = 1081 \ J[/tex]
Part B
[tex]h = 169.7 \ m[/tex]
Explanation:
From the question we are told that
The mass of the blade is [tex]m_b = 0.65 \ kg[/tex]
The length is [tex]l = 0.55 \ m[/tex]
The angular speed is [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]
Generally the moment of inertia of the of this mower is mathematically evaluated as
[tex]I = \frac{m_b * l^2 }{12}[/tex]
substituting values
[tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]
[tex]I = 0.016 \ kg m^2[/tex]
Generally the rotational kinetic energy of the bland is
[tex]R = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]
[tex]R = 1081 \ J[/tex]
At point where the gravitational potential energy is equal to the rotational kinetic energy we have that
[tex]P = R = m_b * h * g[/tex]
Where P is the gravitational potential energy
substituting values
[tex]1081 = 0.65 * 9.8 * h[/tex]
=> [tex]h = 169.7 \ m[/tex]
The “turning effect of a force” (T = F * r) is:
(a) determined as the product of force and the moment of inertia.
(b) generated by concentric forces.
(c) equivalent to the angular momentum.
(d) determined as a product of torque and moment arm.
(e) called “moment” or “torque”.
Answer:
b and e
Explanation:
r x F is the formula for torque.
The "turning effect" or torque happens when concentric forces rotate an object along said center.
a) False because T = Fr = Ia (a = angular acceleration)
b) True
c) False. L = Iw (w = angular velocity), which does not equal Ia
d) False. It is torque, not the product of torque and something else
e) True.
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
4. Mrs. Parker was married to her husband for
30 years. They lived together with their two
children,
(A) Single-parent family
(B) Nuclear family
(C) Blended family
(D) Extended family
I think it’sd
Explanation:
The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids
A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)
(a) 1.00 cm
What is the general expression for the electric field along the axis of a uniformly charged ring? i MN/C
(b) 5.00 cm
i MN/C
(c) 30.0 cm
i MN/C
(d) 100 cm
i MN/C
Answer:
General Expression: E = kql/(l² + r²)^(3/2)
(a) 6.3 MN/C
(b) 22.8 MN/C
(c) 6.1 MN/C
(d) 0.63 MN/C
Explanation:
The general expression for electric field along axis of a uniformly charged ring is:
E = kqL/(L² + r²)^(3/2)
where,
E = Electric Field Strength = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q = Total Charge = 71 μC = 71 x 10⁻⁶ C
L = Distance from center on axis
r = radius of ring = 10 cm = 0.1 m
(a)
L = 1 cm = 0.01 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)
E = (6390 N.m³/C)/(0.00101 m³)
E = 6.3 x 10⁶ N/C = 6.3 MN/C
(b)
L = 5 cm = 0.05 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)
E = (31950 N.m³/C)/(0.00139 m³)
E = 22.8 x 10⁶ N/C = 27.4 MN/C
(c)
L = 30 cm = 0.3 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)
E = (191700 N.m³/C)/(0.03162 m³)
E = 6.1 x 10⁶ N/C = 6.1 MN/C
(d)
L = 100 cm = 1 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)
E = (639000 N.m³/C)/(1.015 m³)
E = 0.63 x 10⁶ N/C = 0.63 MN/C
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.0 kg man just before contact with the ground has a speed of 6.46 m/s. In a stiff-legged landing he comes to a halt in 2.07 ms. Calculate the average net force that acts on him during this time
Answer:
Explanation:
The man comes to halt due to reaction force acting on him in opposite direction . If R be the reaction force
impulse by net force = change in momentum
Net force = R - mg , mg is weight of the man .
( R-mg ) x 2. 07 x 10⁻³ = 73 x 6.46 - 0
R - mg = 227.81 x 10³
Average net force = 227.81 x 10³ N .
When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F
Answer:
B. F/2
Explanation:
The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave
i.e Prad = u
For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.
i.e Prad = 2u.
Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2
5.
The solar system coalesced due to rotational forces and
gravity.
heat.
radioactivity.
solar wind.
Answer:
Gravity
Explanation:
The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.
Cheers.
A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.
Answer:
a) -5.26 m/s
b) 27.91 m
Explanation:
a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.
average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s
The change in velocity in the first half of the interval is ...
Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s
So, the initial velocity (at the beginning of the last 1.85 s interval) is ...
v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)
v1 = -5.259 m/s
__
b) The velocity when the object hits the ground is ...
v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s
This is related to the distance traveled by ...
v² = 2dg . . . . . where g is the acceleration and d is the distance traveled
d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters
The object travels a total distance of about 27.911 meters.
_____
The attached graph shows height vs. time.
In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.
A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?
Answer:
A. F = 0.06 N
B. t = 6.37 s
Explanation:
A)
First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
where,
s = distance traveled = 1.62 m
Vi = 0 m/s (Since, it starts from rest)
t = Time Taken = 4.34 s
a = acceleration = ?
Therefore,
1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²
1.62 m/9.4178 s² = a
a = 0.172 m/s²
Now, from Newton's Second law, we know that:
F = ma
where,
F = Net Force of the combination = ?
m = Mass pf combination = 354 g = 0.354 kg
Therefore,
F = (0.354 kg)(0.172 m/s²)
F = 0.06 N
B)
Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:
F = ma
a = F/m
a = 0.06 N/0.762 kg
a = 0.08 m/s²
Now, using 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²
t² = 1.62 m/(0.04 m/s²)
t = √40.54 s²
t = 6.37 s
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
E = -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E = -4556.18 N/m
The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.
What is electric field?The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.
The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)
The horizontal distance traveled by particle is
x = ucosθt
t = x/ucosθ
The force in an electric field is F = qE...................(1)
where, q is charge , E is the strength of electric field
From, newton 2nd law of motion, Force F = ma.................(2)
Equating both the equations, we get
ma = qE
a = qE/m..................(3)
The vertical distance, y =usinθt - 1/2at²
From equation 3, we have
y = usinθt - 1/2 (qE/m) t²
if y = 0, t = 2musinθ/(qE) = x / (ucosθ)
The electric field is represented as
Also, E = 2mu²×sinθ×cosθ/(xq)
Plug the values, we get
E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)
E = -4556.18 N/m
Thus, the electric field of the particle is -4556.18 N/m.
Learn more about electric field.
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The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
A student is given a small object that is hanging from a ring stand on a nylon thread. The student attempts to charge the object electrically in several ways. Based upon his results, he concludes the object is made of an insulating material. Which set of results must he have collected?
A. The object could be charged only by contact.
B. The object could be charged by either contact or induction.
C. The object could be charged by either contact or polarization.
D. The object could be charged only by polarization.
Answer:(a)
Explanation:
Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.
When the play button is pressed, a CD accelerates uniformly from rest to 430 rev/min in 4.0 revolutions. If the CD has a radius of 7.0 cm and a mass of 17 g , what is the torque exerted on it?
Answer:
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Explanation:
As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.
[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.
The angular acceleration is cleared and calculated:
[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]
Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:
[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]
[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]
The angular accelaration measured in radians per square second is:
[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]
[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]
Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:
[tex]\tau = I \cdot \alpha[/tex]
Where:
[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
In addition, a CD has a form of a uniform disk, whose moment of inertia is:
[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]
Where:
[tex]m[/tex] - Mass of the CD, measured in kilograms.
[tex]r[/tex] - Radius of the CD, measured in meters.
If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:
[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]
[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]
Now, the net torque exerted on CD is:
[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Four long wires are each carrying 6.0 A. The wires are located
at the 4 corners of a square with side length 9.0 cm. All of
these wires are carrying current out of the page. The
magnetic field (in T) at one corner of the square is:
Answer:
[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:
[tex]B=\frac{\mu_oI}{2\pi r}[/tex]
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m
[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]
Then you have:
[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?
Answer:
a. about 20 years younger
b. Malcolm sits around for 49.94 years
c. 2.268x[tex]10^{17}[/tex] m
Explanation:
light travels 3x[tex]10^{8}[/tex] m in one seconds
in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m
for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex] = 3.78x[tex]10^{17}[/tex] m
Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex] = 2.4x[tex]10^{8}[/tex] m/s
time that will pass for Malcolm will be distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]
= 1575000000 s = 49.94 years
the relativistic time t' will be
t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]
t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo
this means that Flo will be about 20 years younger than Malcolm when she returns
relativistic distance is
d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x 0.6
d' = 2.268x[tex]10^{17}[/tex] m this is how far it is to Flo
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..
Answer:
* air resistance.
*the direction of the rotation of the Earth
rotation of the thrown body
Explanation:
The projectile launch is described by the expressions
x-axis x = v₀ₓ t
y-axis y = [tex]v_{oy}[/tex] t - ½ gt²
When the things that affect this movement are analyzed, in order of importance we have:
* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.
* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.
* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption
* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height
A light wave will *Blank* if it enters a new medium perpendicular to the surface.
Answer:
A light wave will not stop if it enters a new medium perpendicular to the surface.
Explanation:
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
What is meant by refraction ?Refraction is defined as an optical phenomenon by which the direction of a light wave gets changed when it travels from one medium to another. This is because of the change in speed.
Here,
The light wave is entering a new medium such that it enters perpendicular to the surface. Angle of incidence is the angle between the incident ray and the line perpendicular to the surface at the point of incidence. Since, here the light ray is incident normal to the surface that means the angle of incidence is 0.
According to Snell's law,
sin i = μ sin r
where i is the angle of incidence, r is the angle of refraction and μ is the constant called refractive index.
As i = 0, sin i = 0
So, μ sin r = 0
Since μ is a constant, we can say that sin r = 0 or the angle of refraction,
r = 0
This means that there is no refraction and hence the light wave won't get deviated when it enters the medium normally.
Hence,
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
To learn more about refraction, click:
https://brainly.com/question/14397443
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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 3.5 mV/m. At what rate is the magnetic field changing?
Answer
The rate at which the magnetic field is changing is [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
Explanation
From the question we are told that
The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]
The radius is [tex]r = 1.5 \ m[/tex]
The rate of change of the magnetic field is mathematically represented as
[tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]
Where [tex]dl[/tex] is change of a unit length
[tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]
Where A is the area which is mathematically represented as
[tex]A = \pi r^2[/tex]
So
[tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]
[tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]
where L is the circumference of the circle which is mathematically represented as
[tex]L = 2 \pi r[/tex]
So
[tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]
[tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]
[tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]
substituting values
[tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]
[tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]
You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied
Answer:
18 N
Explanation:
Force can be found using the following formula.
f= m*a
where m is the mass and a is the acceleration.
We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.
m= 36 kg
a= 0.5 m/s^2
Substitute these values into the formula.
f= 36 kg * 0.5 m/s^2
Multiply 36 and 0.5
f=18 kg m/s^2
1 kg m/s^2 is equivalent to 1 Newton, or N.
f= 18 Newtons
The force being applied is 18 kg m/s^2, Newtons, or N
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
A. 26
B. 35
C. 42
D. 22
E. 48
Answer:
35Explanation:
According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.
sini/sinr = n
n is the constant = refractive index
Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng
nw is the refractive index of water and ng is that of glass
sini/sinr = nw/ng
given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass
On substitution;
sin 30/sinr = 1.33/1.5
1.5sin30 = 1.33sinr
sinr = 1.5sin30/1.33
sinr = 0.75/1.33
sinr = 0.5639
r = arcsin0.5639
r ≈35°
angle the light leave the glass is 35°
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
[tex]^{\circ}C=K-273[/tex]
Put T = 100 K
[tex]T=100-273=-173^{\circ} C[/tex]
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.