Answer:
3.958 × 10²⁶ watt
Explanation:
Given:
Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m
Total solar irradiance at earth orbit = 1400 watt/m²
Now,
Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]
⇒ [tex]A_e[/tex] = [tex]4\pi \times (150\times 10^{8})^2[/tex]
⇒ [tex]A_e[/tex] = [tex]2.827\times 10^{23}m^2[/tex]
Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]
the flux = [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]
⇒the flux = 3.958 × 10²⁶ watt
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.5 cm, and the electric field within the capacitor has a magnitude of 2.5 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?
Answer:
6 x 10⁻¹⁵ J
Explanation:
d = distance between the plates = 1.5 cm = 0.015 m
E = magnitude of electric field between the plates of the capacitor = 2.5 x 10⁶ V/m
q = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C
Force on the electron due to electric field is given as
F = q E
F = (1.6 x 10⁻¹⁹) (2.5 x 10⁶)
F = 4 x 10⁻¹³ N
KE₀ = initial kinetic energy of electron at negative plate = 0 J
KE = final kinetic energy of electron at positive plate = ?
Using work-change in kinetic energy
F d = KE - KE₀
(4 x 10⁻¹³) (0.015) = KE - 0
KE = 6 x 10⁻¹⁵ J
An electron released from rest towards the positive plate of a capacitor, which is separated by a distance of 1.5 cm and an electric field of 2.5 x 10^6 V/m, attains a kinetic energy of 37.5 keV when it reaches the positive plate.
Explanation:The subject you're studying is called electrical potential energy, specifically its conversion into kinetic energy in the context of a parallel plate capacitor. In the case of an electron released from rest towards the positive plate of a capacitor, it is said to be moving through an electrical potential difference. This difference, coupled with the charge of the electron, provides the electron with energy, accelerating it.
Given that the electric field (E) is 2.5 x 106 V/m and the distance between the plates (d) is 1.5 cm or 0.015 m, we can use the formula E = V/d to calculate the potential difference (V). Substituting the given values, the potential difference is 2.5 x 106 V/m * 0.015 m = 37,500 V or 37.5 kV.
Furthermore, as per the relation that an electron accelerated through a potential difference of 1 V attains an energy of 1 electron-volt (eV), an acceleration through 37.5 kV will grant an energy of 37.5 keV. Since its initial kinetic energy was zero (as it was released from rest), this 37.5 keV is the kinetic energy of the electron just as it reaches the positive plate.
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Which of the following systems has constant kinetic and potential energies? A car moving along a level road at constant speed
A car moving up a hill at constant speed
A car moving down a hill at constant speed
All of the above
A parachutist bails out and freely falls 50 m. Then the parachute opens, and thereafter she deceler- ates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?
Answer:
a)The parachutist in the air for 12.63 seconds.
b)The parachutist falls from a height of 293 meter.
Explanation:
Vertical motion of parachutist:
Initial speed, u = 0m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 50 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 50
v = 31.32 m/s
Time taken for this
31.32 = 0 + 9.81 x t
t = 3.19 s
After 50m we have
Initial speed, u = 31.32m/s
Acceleration, a = -2 m/s²
Final speed , v = 3 m/s
We have equation of motion, v² = u² + 2as
Substituting
3² = 31.32² - 2 x 2 x s
s = 243 m
Time taken for this
3 = 31.32 - 2 x t
t = 9.44 s
a) Total time = 3.19 + 9.44 = 12.63 s
The parachutist in the air for 12.63 seconds.
b) Total height = 50 + 243 = 293 m
The parachutist falls from a height of 293 meter.
A solenoid has 332 turns and a length of 14 cm. If a current of 0.88 A produces a magnetic flux density of 0.28 T in the core of the solenoid, what is the relative permeability of the core material?
Answer:
106.83
Explanation:
N = 332, l = 14 cm = 0.14 m, i = 0.88 A, B = 0.28 T
Let ur be the relative permeability
B = u0 x ur x n x i
0.28 = 4 x 3.14 x 10^-7 x ur x 332 x 0.88 / 0.14 ( n = N / l)
ur = 106.83
To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
Answer:
4.4 square meters = 47 square foot
Explanation:
We have
1 meter = 3.28084 foot
1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot
4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot
4.4 square meters = 47 square foot
A uniform steel plate has an area of 0.819 m2. When subjected to a temperature difference between its sides, a heat current* of 31700 W is found to flow through it. What is the temperature gradient? What is the temperature difference when the plate is 0.0475 m thick? The thermal conductivity of steel is 50.2 W/(m·K).
Answer:
ΔT / Δx = 771 K/m
ΔT = 771 x 0.0475 = 36.62 k
Explanation:
P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k
Use the formula of conduction of heat
H / t = K A x ΔT / Δx
So, ΔT / Δx = P / K A
ΔT / Δx = 31700 / (50.2 x 0.819)
ΔT / Δx = 771 K/m
Now
ΔT = 771 x 0.0475 = 36.62 k
In the pair of supply and demand equations below, where x represents the quantity demanded in units of a thousand and p the unit price in dollars, find the equilibrium quantity and the equilibrium price. p = −x2 − 3x + 80 and p = 7x + 5
In this exercise we have to use the knowledge of finance to calculate the equilibrium value and the quantity, so we have:
Equilibrium quantity = 5Equilibrium price = 40Organizing the information given in the statement we have:
p = -x²-3x+80p = 7x+5So equating the two given equations we have:
[tex]-x^2 - 3x + 80 = 7x+5\\x^2 +3x + 7x + 5 - 80 = 0\\x^2 + 10x - 75 = 0\\x^2- 5x + 15x -75 = 0\\x(x-5) + 15(x-5) = 0[/tex]
So we can see that the roots will be x = 5 and x = -15 since the quantity cannot be in negative therefore, the equilibrium quantity will be = 5 So replace that value at:
[tex]p = -(5)^2-3(5) + 80 = 40\\p = 7(5) + 5 = 40[/tex]
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If a copper wire has a resistance of 23.7 Ω at a temperature of 20.3 oC, what resistance does it have at 79.0 oC? (Neglect any change in length or cross-sectional area resulting from the change in temperature.)
Answer:
[tex]R_{79} = 28.91 OHM[/tex]
Explanation:
Resistance ca be determine by using following formula
[tex]R = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
where
R = Conductor resistance = 23.7Ω
Rref = conductor reistance at reference temperature,
α = temperature coefficient of resistance for material, for copper 40.41*10^{-4}
T = Conductor temperature in Celcius = 20.3°C
Tref = reference temperature at which α is specified.
[tex]23.7 = R_{ref}\left [ 1+ 40.41*10^{-4} (20.3) \right ][/tex][tex]R_{ref} =21.92 OHM[/tex]
now for 79 degree celcius
[tex]R_{79} = R_{ref}\left [ 1+ \alpha (T - T_{ref}) \right ][/tex]
[tex]R_{79} =21.92\left [ 1+ 40.41*10^{-4} (79) \right ][/tex]
[tex]R_{79} = 28.91 OHM[/tex]
A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Answer:
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
Explanation:
Force per unit length between two current carrying wires is given by the formula
[tex]F = \frac{\mu_0 i_1 i_2}{2 \pi d}[/tex]
here we know that
[tex]F = 7.83 \times 10 N/m[/tex]
[tex]d = 7.0 cm = 0.07 m[/tex]
[tex]i_1 = 28 A[/tex]
now we will have
[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]
[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]
[tex]i_2 = 978750 A[/tex]
Since the force between wires is attraction type of force so current must be flowing in upward direction
A falling baseball has an acceleration of magnitude 2.1 m/s2. What is its acceleration in feet per second squared?
Answer:
6.889 ft/s^2
Explanation:
1 m = 3.28 feet
So, 2.1 m/s^2 = 2.1 × 3.28 ft/s^2
= 6.889 ft/s^2
In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground. How far does the shot travel?
Answer:
18.7 m
Explanation:
[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s
θ = angle of launch from the horizontal = 47 deg
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 1.80 m
t = time of travel
Using the kinematics equation
[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]
- 1.80 = (9.5) t + (0.5) (- 9.8) t²
t = 2.11 s
Consider the motion along the horizontal direction
x = horizontal displacement of the shot
[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.11 s
Using the kinematics equation
[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]
x = (8.87) (2.11) + (0.5) (0) (2.11)²
x = 18.7 m
The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
What is projectile motion?A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.
Given to us
The initial velocity of the shotput, u = 13 m/s
The angle from the horizontal, θ = 47.0°
The height from which shotput was thrown h = 1.80 m
As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,
[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]
Substitute the values,
[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]
R = 18.7 m
Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
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A diver can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position. If she makes two rotations in 1.5 seconds when in the tuck position, what is her angular speed (rad/sec) when in the straight position?
Answer:
29.3 rad/s
Explanation:
Moment of inertia in straight position, I1 = I
He takes 2 rotations in 1.5 second
So, Time period, T = 1.5 / 2 = 0.75 second
w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s
Moment of inertia in tucked position, I2 = I / 3.5
Let the new angular speed is w2.
By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.
L1 = L2
I1 w1 = I2 w2
I x 8.37 = I / 3.5 x w2
w2 = 29.3 rad/s
To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. The moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. By applying the conservation of angular momentum equation, we can calculate the angular speed in the straight position.
Explanation:To find the angular speed of the diver in the straight position, we can use the conservation of angular momentum. According to the conservation of angular momentum, the product of the initial moment of inertia and initial angular speed is equal to the product of the final moment of inertia and final angular speed.
Let's assume the initial angular speed in the tuck position is denoted by w' and the final angular speed in the straight position is denoted by w. We are given that the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position. Therefore, the final moment of inertia (I) is 3.5 times greater than the initial moment of inertia (I').
The conservation of angular momentum equation can be written as:
I' * w' = I * w
Since the moment of inertia decreases by a factor of 3.5 when changing from the tuck position to the straight position, we have:
w = w' * (I' / I)
Using the given values, we can calculate the angular speed in the straight position.
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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
[tex]U=\dfrac{1}{2}kx^2[/tex]...(I)
Using gravitational potential energy
[tex]U' =mgh[/tex]....(II)
Using energy of conservation
[tex]E_{i}=E_{f}[/tex]
[tex]U_{i}+U'_{i}=U_{f}+U'_{f}[/tex]
[tex]\dfrac{1}{2}kx^2+0=0+mgh[/tex]
[tex]h=\dfrac{kx^2}{2mg}[/tex]
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
[tex]h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}[/tex]
[tex]h=11.653\ m[/tex]
Hence, The maximum height above the point of release is 11.653 m.
Suppose all the mass of the Earth were compacted into a small spherical ball. Part A What radius must the sphere have so that the acceleration due to gravity at the Earth's new surface was equal to the acceleration due to gravity at the surface of the Moon?
Answer:
0.4 times the radius of moon
Explanation:
gravity on moon is equal to the one sixth of gravity on earth.
g' = g / 6
where, g' is the gravity on moon and g be the gravity on earth.
As the earth shrinks, the mass of earth remains same.
The acceleration due to gravity is inversely proportional to the square of radius of planet.
g' ∝ 1/R'² .....(1)
Where, R' is the radius of moon.
g ∝ 1/R² ..... (2)
Where, R be the radius of earth.
Divide equation (1) by (2)
g / g' = R'² / R²
Put g' = g / 6
6 = R'² / R²
2.5 = R' / R
R = R' / 2.5 = 0.4 R'
Thus, the radius of earth should be 0.4 times the radius of moon.
A single conducting loop of wire has an area of 7.26E-2 m2 and a resistance of 117 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.289 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.367 A?
Answer:
[tex]\frac{dB}{dt}[/tex] = 591.45 T/s
Explanation:
i = induced current in the loop = 0.367 A
R = Resistance of the loop = 117 Ω
E = Induced voltage
Induced voltage is given as
E = i R
E = (0.367) (117)
E = 42.939 volts
[tex]\frac{dB}{dt}[/tex] = rate of change of magnetic field
A = area of loop = 7.26 x 10⁻² m²
Induced emf is given as
[tex]E = A\frac{dB}{dt}[/tex]
[tex]42.939 = (7.26\times 10^{-2})\frac{dB}{dt}[/tex]
[tex]\frac{dB}{dt}[/tex] = 591.45 T/s
A certain part of an aircraft engine has a volume of 1.4 ft^3. (a) Find the weight of the piece when it is made of lead. (b) If the same piece is made of aluminum, what is its weight? Determine how much weight is saved by using aluminum instead of lead.
Final answer:
The weight of the piece when made of lead is 992.6 lb. The weight of the piece when made of aluminum is 236.6 lb. By using aluminum instead of lead, a weight of 756 lb is saved.
Explanation:
(a) Weight of the piece when it is made of lead:
To find the weight of the piece when it is made of lead, we need to know the density of lead. The density of lead is approximately 709 lb/ft³. We can use the formula:
Weight = Density x Volume
Given that the volume of the piece is 1.4 ft³, the weight of the piece when it is made of lead can be calculated as:
Weight = 709 lb/ft³ x 1.4 ft³ = 992.6 lb
(b) Weight of the piece when made of aluminum, and weight saved:
To find the weight of the piece when it is made of aluminum, we need to know the density of aluminum. The density of aluminum is approximately 169 lb/ft³. Using the same formula as before:
Weight = Density x Volume
Given that the volume of the piece is still 1.4 ft³, the weight of the piece when it is made of aluminum can be calculated as:
Weight = 169 lb/ft³ x 1.4 ft³ = 236.6 lb
The weight saved by using aluminum instead of lead can be determined by subtracting the weight of the aluminum piece from the weight of the lead piece:
Weight saved = Weight of lead piece - Weight of aluminum piece
Weight saved = 992.6 lb - 236.6 lb = 756 lb
24. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?
Answer:
0.273 m/s
Explanation:
Momentum is conserved:
mu = (m + M) v
(0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v
v = 0.273 m/s
The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.The final velocity will be 0.273 m/sec.
What is law of conservation of momentum?The momentum of the conservation said that the momentum before collision is equal to the momentum before collision.
The given data in the problem is;
m is the mass of clay model before collision= 0.200 Kg
u is the sliding velocity before collision=0.75 m/sec
M is the mass after collision= 0.350 Kg
v is the velocity after collision=?
On applying the law of conservation of momentum we get;
[tex]\rm mu = (m + M) v \\\\\ (0.200 kg) (0.750 m/s) = (0.200 kg + 0.350 kg) v \\\\ v = 0.273 m/s[/tex]
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Convert: Thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C). Surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m20C)
To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), multiply by 694.7. To convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), multiply by 11,545.
Explanation:To convert the thermal conductivity value of 0.3 Btu/(h ft°F) to W/(m °C), we can use the following conversion factors:
1 Btu = 1055.06 J1 ft = 0.3048 m1 °F = 0.5556 °CUsing these conversion factors, we can convert the thermal conductivity value as follows:
0.3 Btu/(h ft°F) = (0.3 * 1055.06 J)/(h * 0.3048 m * 0.5556 °C) = 694.7 W/(m °C)
Similarly, to convert the surface heat transfer coefficient value of 105 Btu/(h ft2 oF) to W/(m2 °C), we can use the following conversion factors:
1 ft2 = 0.0929 m21 °F = 0.5556 °CUsing these conversion factors, we can convert the surface heat transfer coefficient value as follows:
105 Btu/(h ft2 oF) = (105 * 1055.06 J)/(h * 0.0929 m2 * 0.5556 °C) = 11,545 W/(m2 °C)
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The conversion of the thermal conductivity value yields approximately 0.52 W/(m°C), and the surface heat transfer coefficient conversion yields approximately 596.15 W/(m²°C). Conversion factors for BTU to J, ft to m, and °F to °C are critical in these calculations.
To convert thermal conductivity from BTU/(h ft°F) to W/(m°C):
Given value: 0.3 BTU/(h ft°F).Use conversion factors: 1 BTU = 1055.06 J, 1 ft = 0.3048 m, 1 °F = (5/9) °C.Calculation: 0.3 BTU/(h ft°F) * (1055.06 J/BTU) * (1/3600 s/h) * (1/0.3048 m) * (9/5 °C/°F).Simplify: 0.3 * (1055.06/3600) * (1/0.3048) * (9/5) ≈ 0.52 W/(m°C).To convert surface heat transfer coefficient from BTU/(h ft²°F) to W/(m²°C):
Given value: 105 BTU/(h ft²°F).Use conversion factors: 1 BTU = 1055.06 J, 1 ft = 0.3048 m, 1 °F = (5/9) °C.Calculation: 105 BTU/(h ft²°F) * (1055.06 J/BTU) * (1/3600 s/h) * (1/0.3048² m²) * (9/5 °C/°F).Simplify: 105 * (1055.06/3600) * (1/0.3048²) * (9/5) ≈ 596.15 W/(m²°C).In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0265 kg and the mass of the pendulum is 0.250 kg, how high will the pendulum swing if the marble has an initial speed of 5.05 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
0.012 m
Explanation:
m = mass of the marble = 0.0265 kg
M = mass of the pendulum = 0.250 kg
v = initial velocity of the marble before collision = 5.05 m/s
V = final velocity of marble-pendulum combination after the collision = ?
using conservation of momentum
m v = (m + M) V
(0.0265) (5.05) = (0.0265 + 0.250) V
V = 0.484 m/s
h = height gained by the marble-pendulum combination
Using conservation of energy
Potential energy gained by the combination = Kinetic energy of the combination just after collision
(m + M) gh = (0.5) (m + M) V²
gh = (0.5) V²
(9.8) h = (0.5) (0.484)²
h = 0.012 m
Using the concept of inelastic collison and principle of conservation of momentum, the height of swing made by the pendulum would be 0.012 meters
Collison is inelastic :
m1v1 = (m1 + m2)vv = final velocity after the collision(0.0265 × 5.05) = (0.0265 + 0.250)v
0.133825 = 0.2765v
v = (0.133825 ÷ 0.2765)
v = 0.484 m/s
Final velocity after collision = 0.484 m/s
Assuming linear momentum is conserved :
Kinetic energy = Potential Energy0.5mv² = mgh
Mass cancels out
0.5v² = gh
0.5(0.484)² = 9.8h
9.8h = 0.1171
h = (0.1171) ÷ 9.8
h = 0.0119
Therefore, height of the swing made by the pendulum would be 0.012 meters.
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A square, single-turn coil 0.132 m on a side is placed with its plane perpendicular to a constant magnetic field. An emf of 27.1 mV is induced in the winding while the area of the coil decreases at a rate of 0.0785 m2 /s. What is the magnitude of the magnetic field? Answer in units of T.
Answer:
0.35 T
Explanation:
Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s
Use the Faraday's law of electromagnetic induction
e = rate of change of magnetic flux
Let b be the strength of magnetic field.
e = dФ / dt
e = d ( B A) / dt
e = B x dA / dt
0.0271 = B x 0.0785
B = 0.35 T
The magnitude of the magnetic field is 3.45 × 10^-4 T.
Explanation:The magnitude of the induced emf in a square, single-turn coil can be calculated using the equation: emf = -N * A * (d(B)/dt), where N is the number of turns, A is the area of the coil, and d(B)/dt is the rate of change of the magnetic field. In this case, the area of the coil is decreasing at a rate of 0.0785 m2/s, and the emf is given as 27.1 mV. We can rearrange the equation to solve for the magnitude of the magnetic field (B):
B = -emf / (N * d(A)/dt) = -27.1 mV / (1 * 0.0785 m2/s) = -345.222 T/s = 3.45 × 10^-4 T/s.
Therefore, the magnitude of the magnetic field is 3.45 × 10^-4 T.
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A particle with an initial velocity of 50 m slows at a constant acceleration to 20 ms-1 over a distance of 105 m. How long does it take for the particle to slow down? (a) 2 s (c) 3 s (b)4 s (d)5 s
Answer:
Time taken, t = 3 s
Explanation:
It is given that,
Initial velocity of the particle, u = 50 m/s
Final velocity, v = 20 m/s
Distance covered, s = 105 m
Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}[/tex]
[tex]a=-10\ m/s^2[/tex]
So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}[/tex]
t = 3 s
So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.
A small object of mass 3.66g and charge-19 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What ares the magnitude and direction of the electric field? magnitude N/C direction
Explanation:
It is given that,
Mass of the object, m = 3.66 kg
Charge, q = -19 μC = -19 × 10⁻⁶ C
It is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground such that,
[tex]F_g=F_e[/tex]
[tex]F_g\ and\ F_e[/tex] are gravitational and electrostatic forces respectively
[tex]mg=qE[/tex]
[tex]E=\dfrac{mg}{q}[/tex]
[tex]E=\dfrac{3.66\ kg\times 9.8\ m/s^2}{-19\times 10^{-6}\ C}[/tex]
E = −1887789.47 N/C
[tex]E=-1.89\times 10^6\ N/C[/tex]
Negative sign shows that the electric field is in the opposite direction of the electric force. Since, the weight of the object is in downward direction and its electric force (which is balancing its weight) is in upward direction. So, we can say that the electric field is in downward direction.
A 2.00-kg block of aluminum at 50.0 °C is dropped into 5.00 kg of water at 20.0 °C. What is the change in entropy during the approach to equilibrium, assuming no heat is exchanged with the environment? The specific heat of aluminum is 0.22 cal/(g∙K).
To calculate the change in entropy during the approach to equilibrium, use the formula ΔS = mcΔT. For the aluminum block, ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum. For the water, ΔS_water = m_water * c_water * ΔT_water. Set ΔT_aluminum equal to ΔT_water and solve for T_water. Add the change in entropy for the aluminum and water to find the total change in entropy. Simplify the equation by converting cal to J and kg to g. Set the total change in entropy equal to zero and solve for T_aluminum to find the initial temperature of the aluminum.
Explanation:To calculate the change in entropy during the approach to equilibrium, we need to use the formula for entropy change:
ΔS = mcΔT
where ΔS is the change in entropy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the aluminum block is dropped into the water, so the final temperature will be the same for both the aluminum and the water. We can calculate the change in entropy for each substance separately and then add them together to find the total change in entropy.
For the aluminum block:
ΔS_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
ΔS_aluminum = 2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)
For the water:
ΔS_water = m_water * c_water * ΔT_water
ΔS_water = 5.00 kg * 1 cal/(g∙K) * (T_water - 20.0 °C)
Since the final temperature is the same for both substances, we can set ΔT_aluminum equal to ΔT_water and solve for T_water:
50.0 °C - T_aluminum = T_water - 20.0 °C
70.0 °C - T_aluminum = T_water
Substituting this value into the equation for ΔS_water:
ΔS_water = 5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C]
Now we can add the change in entropy for the aluminum and water:
ΔS_total = ΔS_aluminum + ΔS_water
ΔS_total = (2.00 kg * 0.22 cal/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 kg * 1 cal/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
To simplify the equation, we can convert cal to J by multiplying by 4.184 and divide both sides by 1000 to convert kg to g:
ΔS_total = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
We can now solve for T_aluminum by setting ΔS_total equal to 0:
0 = (2.00 * 4.184 J/(g∙K) * (50.0 °C - T_aluminum)) + (5.00 * 4.184 J/(g∙K) * [(70.0 °C - T_aluminum) - 20.0 °C])
Simplifying the equation:
0 = (8.368 J/(g∙K) * (50.0 °C - T_aluminum)) + (20.92 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))
Combining like terms:
0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum)) - (104.6 J/(g∙K) * (70.0 °C - T_aluminum))
Simplifying further:
0 = (29.288 J/(g∙K) * (50.0 °C - T_aluminum) - 104.6 J/(g∙K) * (70.0 °C - T_aluminum)
Expanding the equation:
0 = (29.288 J/(g∙K) * 50.0 °C - 29.288 J/(g∙K) * T_aluminum) - (104.6 J/(g∙K) * 70.0 °C - 104.6 J/(g∙K) * T_aluminum)
Simplifying the equation further:
0 = (1464.4 J - 29.288 J/(g∙K) * T_aluminum) - (7322 J - 104.6 J/(g∙K) * T_aluminum)
Combining like terms:
0 = 1464.4 J - 29.288 J/(g∙K) * T_aluminum - 7322 J + 104.6 J/(g∙K) * T_aluminum
Simplifying the equation:
0 = -5857.6 J + 75.312 J/(g∙K) * T_aluminum
Isolating T_aluminum:
5857.6 J = 75.312 J/(g∙K) * T_aluminum
T_aluminum = (5857.6 J) / (75.312 J/(g∙K))
T_aluminum ≈ 77.7 °C
Therefore, the initial temperature of the aluminum was approximately 77.7 °C.
A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and total charge q. What force does the wire experience?
Answer:
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
Explanation:
As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by
[tex]\lambda = \frac{q}{L}[/tex]
now the electric field due to long line charge at a distance d from it is given as
[tex]E = \frac{2k\lambda}{d}[/tex]
[tex]E = \frac{q}{2\pi \epsilon_0 d}[/tex]
now the force on the other charge in this electric field is given as
[tex]F = QE[/tex]
[tex]F = \frac{Qq}{2\pi \epsilon_0 L d}[/tex]
A parallel plate capacitor of area A = 30 cm2 and separation d = 5 mm is charged by a battery of 60-V. If the air between the plates is replaced by a dielectric of κ = 4 with the battery still connected, then what is the ratio of the initial charge on the plates divided by the final charge on the plates?
Answer:
0.25
Explanation:
A = area of each plate = 30 cm² = 30 x 10⁻⁴ m²
d = separation between the plates = 5 mm = 5 x 10⁻³ m
[tex]C_{air}[/tex] = Capacitance of capacitor when there is air between the plates
k = dielectric constant = 4
[tex]C_{dielectric}[/tex] = Capacitance of capacitor when there is dielectric between the plates
Capacitance of capacitor when there is air between the plates is given as
[tex]C_{air} = \frac{\epsilon _{o}A}{d}[/tex] eq-1
Capacitance of capacitor when there is dielectric between the plates is given as
[tex]C_{dielectric} = \frac{k \epsilon _{o}A}{d}[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{\frac{\epsilon _{o}A}{d}}{\frac{k \epsilon _{o}A}{d}}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{k}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=\frac{1}{4}[/tex]
[tex]\frac{C_{air}}{C_{dielectric}}=0.25[/tex]
Charge stored in the capacitor when there is air is given as
[tex]Q_{air}=C_{air}V[/tex] eq-3
Charge stored in the capacitor when there is dielectric is given as
[tex]Q_{dielectric}=C_{dielectric}V[/tex] eq-4
Dividing eq-3 by eq-4
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}V}{C_{dielectric} V}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=\frac{C_{air}}{C_{dielectric}}[/tex]
[tex]\frac{Q_{air}}{Q_{dielectric}}=0.25[/tex]
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C).1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.34 kg2.34 kg of this oil from 23 °C23 °C to 191 °C?191 °C?
Answer:
Heat energy required = 687.96 kJ
Explanation:
Heat energy required, H = mCΔT.
Mass of cooking oil, m = 2.34 kg = 2340 g
Specific heat of cooking oil, C = 1.75 J/(g⋅°C)
Initial temperature = 23 °C
Final temperature = 191 °C
Change in temperature, ΔT = 191 - 23 = 168 °C
Substituting values
H = mCΔT
H = 2340 x 1.75 x 168 = 687960 J = 687.96 kJ
Heat energy required = 687.96 kJ
The pressure of 4.20 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?
Answer:
6.30 L
Explanation:
P1 = P, V1 = 4.20 L, T1 = T
P2 = P/3, V2 = ?, T2 = T/2
Where, V2 be the final volume.
Use ideal gas equation
[tex]\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}[/tex]
[tex]V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}[/tex]
By substituting the values, we get
V2 = 6.30 L
A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a horizontal frictionless surface. The bullet emerges from the block with a speed of 1.0 km/s. What is the kinetic energy of the block immediately after the bullet emerges? a. 8.0.J 6.4 J 5.3 J 9.4 J 10 J e.
Answer:
6.4 J
Explanation:
m = mass of the bullet = 10 g = 0.010 kg
v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s
v' = final velocity of the bullet after collision = 1 km/s = 1000 m/s
M = mass of the block = 5 kg
V = initial velocity of block before collision = 0 m/s
V' = final velocity of the block after collision = ?
Using conservation of momentum
mv + MV = mv' + MV'
(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'
V' = 1.6 m/s
Kinetic energy of the block after the collision is given as
KE = (0.5) M V'²
KE = (0.5) (5) (1.6)²
KE = 6.4 J
Final answer:
Using the conservation of momentum, and given that the bullet exits the wooden block with a lower speed, we calculate the block's final speed to be 7.5 m/s after the bullet exits.
Explanation:
To solve this problem, we need to apply the law of conservation of momentum. The law states that if no external forces act on a system, the total momentum of the system remains constant, even though individual objects within the system might change velocities. In this case, the system is made up of the bullet and the wooden block. Before the collision, the block is at rest, so its initial momentum is 0. The bullet’s initial momentum is its mass times its velocity (momentum = mass × velocity). After the collision, the bullet exits the block with a lower speed, indicating that some of its momentum has been transferred to the block.
Step 1: Calculate the initial momentum of the bullet:
Initial momentum of the bullet = mass of bullet × initial velocity of bullet
= 0.05 kg × 500 m/s
= 25 kg·m/s
Step 2: Calculate the final momentum of the bullet:
Final momentum of the bullet = mass of bullet × final velocity of bullet
= 0.05 kg × 200 m/s
= 10 kg·m/s
Step 3: Calculate the change in momentum of the bullet:
Change in momentum = Initial momentum - Final momentum
Change in momentum = 25 kg·m/s - 10 kg·m/s
= 15 kg·m/s
Step 4: Calculate the final momentum of the block:
Since the block was initially at rest, its final momentum is equal to the change in momentum of the bullet (due to conservation of momentum). Therefore, the final momentum of the block is 15 kg·m/s.
Step 5: Calculate the final velocity of the block:
Using the final momentum and the mass of the block, we can find the final velocity:
Final velocity of the block = Final momentum of block / mass of block
= 15 kg·m/s / 2 kg
= 7.5 m/s
Thus, the speed of the block after the bullet has come out the other side is 7.5 m/s.
A superconducting solenoid is to be designed to have an interior field of 7.171 T with a maximum current of 1000 A. The permeability of free space is 1.25664 × 10−6 T · m/A. How many windings are required on a 1- meter solenoid length?
Answer:
5706.5
Explanation:
B = 7.171 T, i = 1000 A, μ0 = 1.25664 x 10^-6 T m/A, l = 1 m
Let the number of turns be N.
The magnetic field due to a current carrying solenoid is given by
B = μ0 x N x i / l
B x l / (μ0 x i) = N
N = 7.171 x 1 / (1.25664 x 10^-6 x 1000)
N = 5706.5
To design a superconducting solenoid that generates an interior field of 7.171 T with a maximum current of 1000 A, approximately 5709 turns per meter are required on a 1-meter length of the solenoid.
Explanation:To determine the number of windings required on a 1-meter solenoid to achieve a magnetic field of 7.171 T with a maximum current of 1000 A, we can use the formula for the magnetic field inside a solenoid, which is given by:
B = μ_0 * n * I
Where:
B is the magnetic field inside the solenoidμ_0 is the permeability of free space (μ_0 = 1.25664 × 10⁻⁶ T · m/A)n is the number of turns per meterI is the current through the solenoidRearranging the formula to solve for n, we get:
n = B / (μ_0 * I)
Substituting the given values:
n = 7.171 T / (1.25664 × 10⁻⁶ T · m/A × 1000 A)
Now, calculating the value:
n = 7.171 / (1.25664 × 10⁻⁶ × 1000)
n = 7.171 / (1.25664 × 10⁻³)
n = 5708.57 turns per meter
To achieve the desired magnetic field, the solenoid must have approximately 5709 turns per meter to accommodate the rounding to the nearest whole number, since we can't have a fraction of a turn.
A tennis ball is thrown from a 25 m tall building with a zero initial velocity. At the same moment, another ball is thrown from the ground vertically upward with an initial velocity of 17 m/s. At which height will the two balls meet?
Answer:
The two balls meet in 1.47 sec.
Explanation:
Given that,
Height = 25 m
Initial velocity of ball= 0
Initial velocity of another ball = 17 m/s
We need to calculate the ball
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2+h[/tex]
Where, u = initial velocity
h = height
g = acceleration due to gravity
Put the value in the equation
For first ball
[tex]s_{1}=0-\dfrac{1}{2}gt^2+25[/tex]....(I)
For second ball
[tex]s_{2}=17t-\dfrac{1}{2}gt^2+0[/tex]....(II)
From equation (I) and (II)
[tex]-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0[/tex]
[tex]t=\dfrac{25}{17}[/tex]
[tex]t=1.47\ sec[/tex]
Hence, The two balls meet in 1.47 sec.
The two balls will meet at a height of approximately 16.465 meters above the ground.
To find the height at which the two balls meet, we need to consider their respective equations of motion and solve for the height when they intersect.
For the tennis ball thrown from the building:
[tex]\[ h_1(t) = 25 - \frac{1}{2} g t^2 \][/tex]
For the ball thrown from the ground:
[tex]\[ h_2(t) = 17t - \frac{1}{2} g t^2 \][/tex]
To find the time when they meet, we'll set:
[tex]\[ 25 - \frac{1}{2} g t^2 = 17t - \frac{1}{2} g t^2 \]\[ 25 = 17t \]\[ t = \frac{25}{17} \][/tex]
Now, substitute this value of \( t \) into either equation to find the height at which they meet:
[tex]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \left(\frac{25}{17}\right)^2 \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} g \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 9.8 \times \frac{625}{289} \]\[ h_1\left(\frac{25}{17}\right) = 25 - \frac{1}{2} \times 17.07 \]\[ h_1\left(\frac{25}{17}\right) = 25 - 8.535 \]\[ h_1\left(\frac{25}{17}\right) = 16.465 \][/tex]