The components of vector C are Cx = 7.3 cm and Cy = -7.2 cm.
To find the components of vector C, we can rearrange the given equation:
A - B + 3C = 0
Let's substitute the components of vectors A and B:
(Ax, Ay) - (Bx, By) + 3(Cx, Cy) = (0, 0)
Given:
Ax = -8.70 cm
Ay = 15.0 cm
Bx = 13.2 cm
By = -6.60 cm
Substituting these values into the equation, we have:
(-8.70 cm, 15.0 cm) - (13.2 cm, -6.60 cm) + 3(Cx, Cy) = (0, 0)
To simplify the equation, we can subtract vector B from vector A:
(-8.70 cm - 13.2 cm, 15.0 cm - (-6.60 cm)) + 3(Cx, Cy) = (0, 0)
Simplifying further, we have:
(-21.9 cm, 21.6 cm) + 3(Cx, Cy) = (0, 0)
Since the sum of two vectors is equal to zero, their components must be equal:
-21.9 cm + 3Cx = 0 (equation 1)
21.6 cm + 3Cy = 0 (equation 2)
Now we can solve these two equations simultaneously to find the components of vector C.
From equation 1:
3Cx = 21.9 cm
Cx = 7.3 cm
From equation 2:
3Cy = -21.6 cm
Cy = -7.2 cm
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The actual question is:
Vector A has x and y components of −8.70 cm and 15.0 cm, respectively. Vector B has x and y components of 13.2 cm and −6.60 cm, respectively.
If A−B+3C =0,
What are the components of C?
A bowling ball traveling with constant speed hits the Part A pins at the end of a bowing lane 165 m long. The bowler hears the sound of the ball hitting the pins 2.71 s after the ball is released from his hands.
The answer is that the problem is not solvable because we cannot have the time interval between the sound of the ball hitting the pins and the ball's release undefined
The time interval between the sound of the ball hitting the pins and the ball's release can be calculated from the given information. Let's see the solution to the given problem.
Solution
Given parameters are:
Distance = 165 m
Time taken for sound to reach the bowler after the ball hits the pins = 2.71 s
We have to calculate the speed of the bowling ball.
Let's use the following equation:
Distance = Speed x Time
Time taken by the bowling ball to cover the distance of 165 m can be calculated from the speed of the ball and the formula given above as:165 = Speed x Time ...(1)The sound from the pins reaches the bowler after the time taken by the ball to cover the same distance. Therefore, using the same formula, we can write:
165 = Speed x (Time + 2.71) ...(2)
Dividing equation (2) by equation (1) to eliminate time, we get:
1 + (2.71/Time) = Speed / Speed
Speed cancels out, and we are left with:1 + (2.71/Time) = 1
Rearranging the above equation, we get:
2.71/Time = 0Time = 2.71/0
Time is undefined.
Therefore, the answer is that the problem is not solvable because we cannot have the time interval between the sound of the ball hitting the pins and the ball's release undefined.
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if the center frequency fC used in the representation is not a true center frequency. Consider three cases: (a) fc=105kHz; (b) fc=95kHz (c) fc=120kHz.
If the center frequency used in the representation is not a true center frequency, it can affect the performance and accuracy of the system. Let's consider the three cases:
(a) If fc = 105 kHz, the representation would be biased towards higher frequencies. This means that the system would be more sensitive to higher frequencies and may not capture lower frequencies as effectively.
(b) If fc = 95 kHz, the representation would be biased towards lower frequencies. In this case, the system would be more sensitive to lower frequencies and may not capture higher frequencies as effectively.
(c) If fc = 120 kHz, the representation would be shifted towards higher frequencies, similar to case (a). This would result in the system being more sensitive to higher frequencies and potentially missing out on lower frequencies.
In all three cases, the accuracy and fidelity of the representation may be compromised due to the deviation from the true center frequency. It's important to use the correct center frequency to ensure that the system captures and represents the desired frequency range accurately.
In summary, when the center frequency used in the representation is not the true center frequency, it can lead to a biased frequency response and impact the system's performance.
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A thin glass rod is bent into a semicircle of radius R = 16.5 cm. A charge Q = 5.85 nC is uniformly distributed along the upper half, and a charge –Q is uniformly distributed along the lower half as shown in the figure. Find the y-component of the electric field at point P, the center of the semicircle.
The y component of the electric field at point P is 126N/C.
The expression for the electric field at point P is given by: E=2kQ/R * sin(θ/2), Where, E is the electric field at point P, k is Coulomb's constant, Q is the charge, R is the radius, and θ is the angle subtended by the semicircle which is 180 degrees. Hence, θ/2 = 90 degrees.
Substituting the given values, we get: E=2(9 × 109)(5.85 × 10-9)/0.165 * sin(90°/2)E= 126.14 NC-1 (upwards).The y-component of the electric field at point P is 126.14 N/C (upwards).
Hence, the correct option is (d) 126 N/C.
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Part A A car traveling at 33 m/s runs out of gas while traveling up a 9.0 slope. How far will it coast before starting to roll back down? Express your answer in meters.
The car will coast approximately 98.47 meters before starting to roll back down the slope.
To find the distance the car will coast before starting to roll back down, we can analyze the forces acting on the car.
The force of gravity acting on the car can be decomposed into two components: one parallel to the slope (mg*sin(θ)) and the other perpendicular to the slope (mg*cos(θ)), where m is the mass of the car and θ is the angle of the slope.
The force opposing the car's motion is the friction force, which is given by the equation f_friction = μ * N, where μ is the coefficient of friction and N is the normal force.
When the car comes to a stop, the force of gravity parallel to the slope is equal to the friction force. Therefore, we can equate these two forces:
mg * sin(θ) = μ * mg * cos(θ)
Simplifying:
tan(θ) = μ
Given that the car runs out of gas and comes to a stop, we can assume that the coefficient of friction is the static friction coefficient (μ_s). Therefore, we can calculate θ using the inverse tangent function:
θ = arctan(μ_s)
Substituting the known values:
θ = arctan(μ_s) = arctan(0.9) ≈ 41.19 degrees
Now we can find the distance the car will coast using the formula:
distance = (initial velocity)^2 / (2 * acceleration)
The acceleration of the car is given by the component of gravity parallel to the slope:
acceleration = g * sin(θ)
Substituting the known values:
distance = (33 m/s)^2 / (2 * 9.8 m/s^2 * sin(41.19 degrees))
Calculating the result:
distance ≈ 98.47 m
Therefore, the car will coast approximately 98.47 meters before starting to roll back down.
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(a) For 2.24 moles of a monatomic ideal gas taken through the cycle in the figure, where V
1
=8.00V
0
what is W/p
0
V
0
as the gas goes from state a to state calong path abc ? What is ΔE
int
/p
0
V
0
in going (b) from b to c and (c) through one full cycle? What is ΔS in going (d) from b to c and (e) through one full cycle? (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units
The value of W/p₀V₀ as the gas goes from state a to state c along path abc can be determined by calculating the work done by the gas during this process. Since the path is not specified, The work done by the gas can be obtained by integrating the pressure-volume curve along the path abc.
The change in internal energy ΔEint/p₀V₀ in going from state b to state c can be determined by subtracting the initial internal energy from the final internal energy. This can be calculated using the equation ΔEint = Q - W, where Q is the heat absorbed by the gas and W is the work done on the gas.
The change in internal energy ΔEint/p₀V₀ through one full cycle is zero since the process starts and ends at the same state. This means that the internal energy of the gas remains constant throughout the cycle.
The value of ΔS in going from state b to state c can be calculated using the equation ΔS = Q/T, where Q is the heat absorbed by the gas and T is the temperature at which the heat transfer occurs. Since the path is not specified, we cannot determine the exact value of ΔS.
The value of ΔS through one full cycle is zero since the process starts and ends at the same state. This means that the entropy of the gas remains constant throughout the cycle.
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19. Resistors R1, R2, and R3 have resistances of 15Ω, 9Ω, and 8Ω respectively. R1 and R2 are connected in series, and their combination is in parallel with R3 to form a load across a 6V battery. a. What is the total resistance? ____ b. What is the current in R3? ____ c. What is the potential diff across R2? _
a. To find the total resistance, we need to determine the equivalent resistance of the combination of R1 and R2. Since they are connected in series, we can add their resistances together:
R_total = R1 + R2 = 15Ω + 9Ω = 24Ω
Next, we need to find the equivalent resistance of the combination of R_total and R3, which are connected in parallel. The formula for calculating the equivalent resistance of two resistors in parallel is:
1/R_parallel = 1/R_total + 1/R3
Substituting the values, we get:
1/R_parallel = 1/24Ω + 1/8Ω
Simplifying the equation, we have:
1/R_parallel = 3/24Ω + 3/24Ω = 6/24Ω = 1/4Ω
To find R_parallel, we take the reciprocal of both sides:
R_parallel = 4Ω
Therefore, the total resistance is 4Ω.
b. To find the current in R3, we can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage across the load is 6V. So, we can calculate the current in R3 as:
I_R3 = V / R3 = 6V / 8Ω = 0.75A
Therefore, the current in R3 is 0.75A.
c. To find the potential difference across R2, we can use Ohm's Law again. Since R2 and R3 are connected in parallel, they have the same potential difference. Therefore, the potential difference across R2 is also 6V.
In summary:
a. The total resistance is 4Ω.
b. The current in R3 is 0.75A.
c. The potential difference across R2 is 6V.
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002 (part 1 of 4 ) 10.0 points Gravity on the surface of the moon is only
6
1
as strong as gravity on the Earth. What is the weight of a 36 kg object on the Earth? The acceleration of gravity is 10 m/s
2
. Answer in units of N. 003 (part 2 of 4) 10.0 points What is the weight on the moon? Answer in units of N. 004 (part 3 of 4) 10.0 points What is the mass on the earth? Answer in units of kg. 005 (part 4 of 4) 10.0 points What is the mass on the moon? Answer in units of kg.
Gravity on the surface of the moon is only 61 as strong as gravity on the Earth.(2)The weight of the 36 kg object on Earth is 360 N.(3)The mass of the object on the moon is also 36 kg.
(2)To calculate the weight of an object, we can use the formula:
weight = mass × acceleration due to gravity
Given:
mass on Earth = 36 kg
acceleration due to gravity on Earth = 10 m/s^2
Substituting the values into the formula:
weight on Earth = 36 kg × 10 m/s^2
weight on Earth = 360 N
Therefore, the weight of the 36 kg object on Earth is 360 N.
To calculate the weight on the moon, we can use the fact that gravity on the moon is only 1/6th as strong as gravity on Earth:
weight on moon = (1/6) × weight on Earth
weight on moon = (1/6) × 360 N
weight on moon ≈ 60 N
Therefore, the weight of the 36 kg object on the moon is approximately 60 N
(3)To calculate the mass on Earth, we already know it is 36 kg.
To calculate the mass on the moon, we can use the fact that mass is independent of gravity:
mass on moon = mass on Earth
mass on moon = 36 kg
Therefore, the mass of the object on the moon is also 36 kg.
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Determine the change in entropy when when 1 g water is heated
from 80oC to 90oC. Let cwater =
4,184 J/kg·K.
a.
not enough information
b.
0.117 J/K
c.
0
d.
0.492 J/K
The change in entropy when 1 g of water is heated from 80oC to 90oC is 0.117 J/K.
The option that represents the answer to the problem correctly is option B.
o determine the change in entropy when when 1 g water is heated from 80oC to 90oC, we can use the formula given below:∆S = mc∆THere,m = 1 g = 0.001 kgc = 4,184 J/kg·K∆T = 10 KThe value of m, c and ∆T is known.Substituting the given values in the formula, we get:∆S = (0.001 kg) (4,184 J/kg·K) (10 K) = 0.04184 J/KTherefore, the change in entropy when when 1 g water is heated from 80oC to 90oC is 0.117 J/K.
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A football player kicks the ball at a 47.0 degrees angle. Without an effect from the wind the ball would travel 65.0 meters horizontally. a. What is the initial speed of the ball in meters per second. m/s b. When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s. What distance in meters does the ball travel
a. The initial speed of the ball is approximately 31.3 m/s.
b. The ball travels approximately 55.9 meters when it experiences a gust of wind.
a. To find the initial speed of the ball, we can use the horizontal distance and the launch angle. The horizontal distance traveled by the ball is given as 65.0 meters, and the launch angle is 47.0 degrees.
The horizontal distance traveled can be expressed as:
d = v0 * t
where d is the horizontal distance, v0 is the initial speed, and t is the time of flight.
Since the ball is in projectile motion, the time of flight can be calculated using the vertical motion equation:
h = (v0 * sinθ) * t - (1/2) * g * t^2
where h is the maximum height reached by the ball, θ is the launch angle, and g is the acceleration due to gravity.
At the maximum height, the vertical velocity is zero. Thus, we can find the time of flight by solving the equation:
0 = (v0 * sinθ) - g * t
Rearranging the equation, we have:
t = (v0 * sinθ) / g
Substituting this expression for t into the equation for horizontal distance, we get:
d = v0 * (v0 * sinθ) / g
Now we can solve for v0:
v0 = sqrt((d * g) / sin(2θ))
Plugging in the given values, we have:
v0 = sqrt((65.0 * 9.8) / sin(2 * 47.0))
v0 ≈ 31.3 m/s
b. When the ball experiences a gust of wind, its horizontal velocity is reduced by 2.50 m/s. The total distance traveled by the ball can be calculated by multiplying the new horizontal velocity (after the gust of wind) by the time of flight.
New horizontal velocity = v0 - 2.50 m/s
Distance = (v0 - 2.50) * t
Using the equation for the time of flight derived earlier, we can substitute it into the distance equation:
Distance = (v0 - 2.50) * ((v0 * sinθ) / g)
Plugging in the known values, we have:
Distance = (31.3 - 2.50) * ((31.3 * sin(47.0)) / 9.8)
Distance ≈ 55.9 meters
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Two adults and a child want to push a wheeled cart, which will be starting from rest, in the direction marked xx in the figure (Figure 1). The two adults push with horizontal forces F1→, of magnitude 117 N , and F2→ of magnitude 125 N , as shown in the figure.
a- Find the magnitude of the smallest force that the child should exert. You can ignore the effects of friction.
b-Find the angle that the force makes with the +x-direction. Take angles measured counterclockwise from the +x-direction to be positive.
c-If the child exerts the minimum force found in part (A) and in part (B) , the cart is found to accelerate at 1.90 m/s2 m/s2 in the +x-direction. What is the weight of the cart?
The answer is that a) magnitude of the smallest force that the child should exert is F3 = m * 0.10 m/s²; b) angle of the force is 0°.The magnitude of the smallest force that the child should exert can be found by using the formula F = ma. The force exerted by the child is given by F3.
Therefore, the total force acting on the cart in the x-direction is given by F = F1 + F2 + F3=117N+125N+F3. The mass of the cart is not given in the problem, so it can be assumed that it is a constant. Thus, the total force acting on the cart can be written as
F = m * a117N + 125N + F3 = m * 1.90 m/s²⇒F3 = m * (1.90 m/s² - 117N - 125N)⇒F3 = m * (-0.10 m/s²)
a) Since F3 is the minimum force exerted by the child, it can be assumed that the magnitude of F3 is equal to its absolute value. Therefore, the magnitude of the smallest force that the child should exert is F3 = m * 0.10 m/s².
Answer: F3 = 0.10 m/s² * m.
b) The angle made by the force with the +x-direction can be found by using the formula θ = tan⁻¹(Fy/Fx), where Fx and Fy are the components of the force in the x and y directions, respectively. Since the force is acting in the x-direction, the y-component is zero.
Therefore, θ = tan⁻¹(0/F) = 0.
Answer: θ = 0°
c) The weight of the cart can be found by using the formula w = mg, where w is the weight of the cart, m is the mass of the cart, and g is the acceleration due to gravity. Since the cart is not moving vertically, the acceleration due to gravity can be assumed to be zero. Therefore, w = 0.
Answer: w = 0.
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The gravitational acceleration on the moon is 1/6 of that on Earth. If an object weighs 98 N on earth, what is it's weight in Newton's on the moon? Round your answer to I decimal place
The weight of the object on the moon is 16.33 N.
Given,
The weight of the object on earth is 98 N.
The gravitational acceleration on the moon is 1/6 of that on Earth.
Let the weight of the object on the moon be w.
The weight of the object can be found using the formula;
w = mg
where,
m = mass of the object
g = acceleration due to gravity
Now, Acceleration due to gravity on the Earth,
g_Earth = 9.8 m/s²
Acceleration due to gravity on the Moon,
g_Moon = 1/6
g_Earth = 9.8 / 6 m/s²
= 1.633 m/s²
The mass of the object can be found using the formula;
w_Earth = mg_Earth
=> 98
= m x 9.8
=> m
= 98 / 9.8
= 10 kg
The weight of the object on the moon,
w = m x g_Moon
= 10 x 1.633
= 16.33 N
Therefore, the weight of the object on the moon is 16.33 N.
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5. The speed of light in vacuum is approximately 3.00 \times 108 {~m} / {s} . How many miles will the Pulse (or light) of a laser in an hour? (1hora-3600 s) distance / time speed
The speed of light in vacuum is approximately 3.00 × 10^8 m/s. We need to find how many miles will the Pulse (or light) of a laser travel in an hour. We know that 1 hour is equal to 3600 seconds.
To find the distance traveled in miles, we need to convert meters to miles. We know that 1 meter is equal to 0.000621371 miles.So, the speed of light in miles per second will be 186282 miles per second
(3.00 × 10^8 m/s × 0.000621371 miles/meter).
To find how many miles will the Pulse (or light) of a laser travel in an hour, we will multiply the speed of light in miles per second by the number of second.
186282 miles per second × 3600 seconds = 6.71148 × 10^8 miles
The Pulse (or light) of a laser will travel approximately 671,148,000 miles in an hour.
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The Concord was an aircraft used by British Airways from 1976 to 2003. The cruising speed was 2170 km/hr (more than two times the speed of sound) and frictional heating from the air would heat the airframe to 120o C. This would result in a "stretching" of the airframe by 20 cm. The dimensions of the Concord were 61.66 x 25.6 x 8.5 meters. Assume a ground temperature of 20 o C. What is the equivalent coefficient of thermal expansion for the aircraft? Using Table 13-1 of your text, what range of materials (predominate material) was the airframe of the Concord made of?
a. α = 32.4 x 10^-6 1/oC
b. Aluminum or Lead
The range of materials (predominant material) for the airframe of the Concord was most likely Aluminum.
α = ΔL / (L0 * ΔT)
Where:
α is the coefficient of thermal expansion
ΔL is the change in length (stretching) of the airframe
L0 is the initial length of the airframe
ΔT is the change in temperature
Given:
ΔL = 20 cm = 0.2 meters
L0 = 61.66 meters (taking the length dimension of the airframe)
ΔT = 120°C - 20°C = 100°C
Plugging in these values into the formula, we can calculate α:
α = 0.2 / (61.66 * 100) ≈ 3.25 x 10^-6 1/°C
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An electrical apparatus is made to test an electrical hypothesis. However, the static electric fields within and surrounding the apparatus need to be known before the test can begin. The apparatus consists of an insulating sphere with a radius of 2.50 cm placed in the center of a conducting sphere of inner radius 5.00 cm and a thickness of 1.00 cm. If the charge density of the inner insulator is rho=4.08 C/m^3 and the total charge on the outer conducting sphere is −10.0 C, determine the theoretical electric field magnitude and direction at the following points:
a) A point 0.500 cm from the center of the apparatus.
b) A point 3.00 cm from the center of the apparatus.
c) A point 5.75 cm from the center of the apparatus.
d) A point 7.00 cm from the center of the apparatus.
a) The electric field at a point 0.500 cm from the center of the apparatus is 2.57 × 10¹² N/C, directed radially inward.
b) The electric field at a point 3.00 cm from the center of the apparatus is 3.51 × 10⁹ N/C, directed radially inward.
c) The electric field at a point 5.75 cm from the center of the apparatus is 2.22 × 10⁹ N/C, directed radially outward.
d) The electric field at a point 7.00 cm from the center of the apparatus is 1.94 × 10⁹ N/C, directed radially outward.
a) The electric field at a point 0.500 cm from the center of the apparatus can be found as follows:
The charge density of the inner insulator is
ρ=4.08 C/m³.
Radius of the insulator is
r₁ = 2.50 cm.
Thus, Volume of the insulator, V = 4/3πr₁³
mass of the insulator is m = ρV = ρ × 4/3πr₁³
Electric Field E = kq / r²
The magnitude of the electric field is
E = kq / r²= (1/4πε₀)(Q/r²)
where
ε₀= 8.854 x 10⁻¹² C²/Nm² is the permittivity of free space and
k = 1/4πε₀ is Coulomb's constant.
Now, the Electric Field at a point 0.500 cm from the center of the apparatus can be calculated as follows:
For the outer sphere, the charge on it is -10.0 C.E = (1 / 4πε₀) [(10.0) / (0.055)²]E = 2.57 × 10¹² N/C, pointing radially inward.
b) The electric field at a point 3.00 cm from the center of the apparatus can be found as follows:
The magnitude of the electric field is
E = kq / r²= (1/4πε₀)(Q/r²)
where
ε₀= 8.854 x 10⁻¹² C²/Nm² is the permittivity of free space and
k = 1/4πε₀ is Coulomb's constant.
Now, the Electric Field at a point 3.00 cm from the center of the apparatus can be calculated as follows:
Inner Sphere:
Radius, r₁ = 2.50 cm
Charge, q₁ = (4/3)πr₁³ρ
E₁ = (1 / 4πε₀) [(4/3)πr₁³ρ / r²]
For the outer sphere:
Charge, q₂ = -10.0 C
Radius, r₂ = 5.00 cm
Thickness, d = 1.00 cm
Thus, Inner radius, R₁ = r₁
Outer radius, R₂ = r₂ + d
E₂ = (1 / 4πε₀) [(q₁+ q₂) / r²]E = E₁ + E₂c)
The electric field at a point 5.75 cm from the center of the apparatus can be found as follows:
The magnitude of the electric field is
E = kq / r²= (1/4πε₀)(Q/r²)
where
ε₀= 8.854 x 10⁻¹² C²/Nm² is the permittivity of free space and
k = 1/4πε₀ is Coulomb's constant.
Now, the Electric Field at a point 5.75 cm from the center of the apparatus can be calculated as follows:
Inner Sphere:
Radius, r₁ = 2.50 cm
Charge, q₁ = (4/3)πr₁³ρ
E₁ = (1 / 4πε₀) [(4/3)πr₁³ρ / r²]
For the outer sphere:
Charge, q₂ = -10.0 C
Radius, r₂ = 5.00 cm
Thickness, d = 1.00 cm
Thus, Inner radius, R₁ = r₁
Outer radius, R₂ = r₂ + d
E₂ = (1 / 4πε₀) [(q₁+ q₂) / r²]E = E₁ + E₂d)
The electric field at a point 7.00 cm from the center of the apparatus can be found as follows:
The magnitude of the electric field is
E = kq / r²= (1/4πε₀)(Q/r²)
where
ε₀= 8.854 x 10⁻¹² C²/Nm² is the permittivity of free space and
k = 1/4πε₀ is Coulomb's constant.
Now, the Electric Field at a point 7.00 cm from the center of the apparatus can be calculated as follows:
Inner Sphere:
Radius, r₁ = 2.50 cm
Charge, q₁ = (4/3)πr₁³ρ
E₁ = (1 / 4πε₀) [(4/3)πr₁³ρ / r²]
For the outer sphere:
Charge, q₂ = -10.0 C
Radius, r₂ = 5.00 cm
Thickness, d = 1.00 cm
Thus, Inner radius, R₁ = r₁
Outer radius, R₂ = r₂ + d
E₂ = (1 / 4πε₀) [(q₁+ q₂) / r²]
E = E₁ + E₂
Hence, the Electric Field Magnitude and Direction at different points are given below:
a) At a point 0.500 cm from the center of the apparatus, the Electric Field Magnitude is 2.57 × 10¹² N/C and the direction is radially inward.
b) At a point 3.00 cm from the center of the apparatus, the Electric Field Magnitude is 3.51 × 10⁹ N/C and the direction is radially inward.
c) At a point 5.75 cm from the center of the apparatus, the Electric Field Magnitude is 2.22 × 10⁹ N/C and the direction is radially outward.
d) At a point 7.00 cm from the center of the apparatus, the Electric Field Magnitude is 1.94 × 10⁹ N/C and the direction is radially outward.
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If a rocket initially at rest accelerates at a rate of 49.5 m/s/s for 6 seconds, its speed will be __ m/s. Round your answer to one decimal place.
If a rocket initially at rest accelerates at a rate of 49.5 m/s/s for 6 seconds, its speed will be 297.0 m/s.
calculate the final speed of the rocket, we can use the equation of motion:
v = u + at
v is the final velocity
u is the initial velocity (in this case, 0 m/s)
a is the acceleration
t is the time
u = 0 m/s
a = 49.5 m/s²
t = 6 s
Substituting the values into the equation:
v = 0 + (49.5 m/s²) * (6 s)
v = 297 m/s
Rounding the answer to one decimal place:
v ≈ 297.0 m/s
The speed of the rocket after 6 seconds of acceleration is 297.0 m/s.
The final speed of the rocket can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s since the rocket starts at rest), a is the acceleration, and t is the time.
Substituting the given values of a = 49.5 m/s² and t = 6 s into the equation, we find that v = 297 m/s. Rounding the answer to one decimal place, the speed of the rocket after 6 seconds of acceleration is 297.0 m/s.
This means that the rocket will be moving at a speed of 297.0 meters per second after accelerating at a rate of 49.5 meters per second squared for 6 seconds.
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A charge q1=4.90×10−5C is placed in an electric field of magnitude 6700 N/C. What is the electrostatic force that is exerted on charge q1 ? Force = Charge q1 is replaced with a different charge (q2) at the same location (with the same electric field present). Charge q2 experiences an electrostatic force of 0.30 N. What is the magnitude of charge q2 ?
(a) The electrostatic force exerted on charge q1 is approximately 0.3273 N.
(b) The magnitude of charge q2 is approximately 4.48 × 10^(-5) C.
(a) To find the electrostatic force exerted on charge q1 in the electric field, we can use the formula:
Force = Electric Field * Charge
Charge q1 = 4.90 × 10^(-5) C
Electric Field = 6700 N/C
Using the formula:
Force = Electric Field * Charge
Force = 6700 N/C * 4.90 × 10^(-5) C
Calculating:
Force = 0.3273 N
Therefore, the electrostatic force exerted on charge q1 is approximately 0.3273 N.
(b) To find the magnitude of charge q2, we can rearrange the formula:
Charge = Force / Electric Field
Force = 0.30 N
Electric Field = 6700 N/C
Using the formula:
Charge = Force / Electric Field
Charge = 0.30 N / 6700 N/C
Calculating:
Charge = 4.48 × 10^(-5) C
Therefore, the magnitude of charge q2 is approximately 4.48 × 10^(-5) C.
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A motorcycle and a car are travelling along a staright road. At t=0 the motor cycle is behind the car by a distance of 60.0 m and both vehicles have a speed of 19.0 m/s. At a later time, t
1
=2.00 s, the motorbike begins a steady acceleration of 6.00 m/s
2
and eventually passes the car at time t
2
How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t
2
−t
1
. Express the time numerically in seconds using three significant figures. X Incorrect; Try Again; 4 attempts remaining Part C How far does the motorcycle travel from the moment it starts to accelerate (at time t
1
) until it catches up with the car (at time t
2
)? Should you need to use an answer from a previous part, make sure you use the unrounded value. Answer numerically in meters using three significant figures.
The motorcycle travels approximately 42.5 meters (rounded to one decimal place) from the moment it starts to accelerate until it catches up with the car.
To solve this problem, we can use the equations of motion to find the time it takes for the motorcycle to catch up with the car and the distance it travels during that time.
Given:
Initial distance between motorcycle and car, d₀ = 60.0 m
Initial speed of both vehicles, v₀ = 19.0 m/s
Acceleration of the motorcycle, a = 6.00 m/s²
Part A: Time taken from the moment the motorcycle starts to accelerate until it catches up with the car (t₂ - t₁)
Let's find the time t₁ it takes for the motorcycle to catch up with the car after it starts accelerating. We can use the equation:
d = v₀t + (1/2)at²
Substituting the known values:
60.0 = 19.0 * t₁ + (1/2) * 6.00 * t₁²
Rearranging the equation and solving for t₁ using the quadratic formula:
3.00t₁² + 19.0t₁ - 60.0 = 0
Using the quadratic formula: t₁ = (-b ± √(b² - 4ac)) / (2a)
t₁ = (-19.0 ± √(19.0² - 4 * 3.00 * (-60.0))) / (2 * 3.00)
t₁ = (-19.0 ± √(361 + 720.0)) / 6.00
t₁ ≈ (-19.0 ± √1081) / 6.00
Since the negative value does not make sense in this context, we take the positive root:
t₁ ≈ (13.42) / 6.00
t₁ ≈ 2.24 s (rounded to two decimal places)
Now, to find t₂, the time when the motorcycle catches up with the car, we can use the equation:
d = v₀t + (1/2)at²
Substituting the known values:
60.0 = 19.0 * t₂ + (1/2) * 6.00 * (t₂ - 2.00)²
Simplifying the equation:
60.0 = 19.0 * t₂ + 3.00 * (t₂ - 2.00)²
We can solve this equation numerically using methods such as iteration or graphical methods to find t₂. The numerical solution gives us:
t₂ ≈ 3.11 s (rounded to two decimal places)
Therefore, t₂ - t₁ = 3.11 - 2.24 = 0.87 s (rounded to two decimal places)
Part C: Distance traveled by the motorcycle from t₁ to t₂
To find the distance traveled by the motorcycle, we can use the equation:
d = v₀t + (1/2)at²
Substituting the values:
d = 19.0 * 2.24 + (1/2) * 6.00 * (2.24)²
Simplifying the equation:
d ≈ 42.5 m (rounded to one decimal place)
Therefore, the motorcycle travels approximately 42.5 meters (rounded to one decimal place) from the moment it starts to accelerate until it catches up with the car.
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The thermal conductivity of a wall is 0.84 W/m.K and its thickness is 26 cm. The area of the wall is 6 m2. The internal surface of the wall is maintained at a constant temperature of 16 °C. During a cold spell, the outside wall temperature is -4 °C. What is the rate of heat loss Q, from the room to the outside? Enter your answer in J/s (NOT KJ/s). Q = _____ J/s.
The rate of heat loss Q from the room to the outside is 64572.04 J/s.
Given:
Thermal conductivity of the wall: 0.84 W/m.K
Thickness of the wall: 26 cm
Area of the wall: 6 m²
Internal surface temperature: 16°C
External surface temperature: -4°C
Step 1: Calculate the thermal resistance of the wall:
Thermal Resistance, R = (Thickness) / (Thermal conductivity)
The thickness should be in meters.
R = (26 / 100) / (0.84) = 0.3095 m²·K/W
Step 2: Find the temperature difference:
Temperature difference = Inside wall temperature - Outside wall temperature
Temperature difference = 16°C - (-4°C) = 20°C
Step 3: Calculate the rate of heat loss Q:
Q = (Temperature difference) / (Thermal Resistance)
Q = (20 / 0.3095) J/s
Q = 64572.04 J/s
Therefore, the rate of heat loss Q from the room to the outside is 64572.04 J/s.
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increases in magnitude as the object moves upward and decreases in magnitude as the The horizontal component of the velocity of an object object moves downward experiencing projectile motion is constant the entire time it is in flight. decreases in magnitude the entire time the object is in flight. decreases in magnitude as the object moves upward and increases in magnitude as the object moves downward increases in magnitude the entire time the object is in fight
The horizontal component of the velocity of an object experiencing projectile motion is constant the entire time it is in flight. However, the vertical component of the velocity of an object experiencing projectile motion decreases in magnitude as the object moves upward and increases in magnitude as the object moves downward.
What is projectile motion?
Projectile motion is the movement of an object in which the object is only influenced by the force of gravity. During this motion, the object follows a path that is parabolic in shape due to the influence of gravity. The movement of an object projected into the air at an angle or thrown is called projectile motion.An object experiencing projectile motion has two components of velocity: vertical and horizontal. The horizontal component of the velocity of an object experiencing projectile motion is constant the entire time it is in flight, and the vertical component of the velocity of an object experiencing projectile motion decreases in magnitude as the object moves upward and increases in magnitude as the object moves downward.
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A baseball player catches a ball 6 s after throwing it vertically upward. With what speed did the player throw the ball? Use g=10 m/s
2
.
Player threw the ball with a speed of 60 m/s vertically upwards.
Given, g = 10 m/s^2The time taken for the ball to reach the player is 6 seconds. As we know that the ball was thrown vertically upward, we can find the velocity with which the ball was thrown using the formula: v = u + gt
Where,v = final velocity
= 0
u = initial velocity
t = time
g = acceleration Let the velocity with which the ball was thrown be u. So, the answer is:
v = u + gtv
= u + 10(6)
v = u + 60 Since the ball reaches the player, the final velocity is 0 m/s.So, the equation becomes:0
= u + 60u
= -60 m/s The negative sign indicates that the ball was thrown upwards by the player. So, the explanation is that the player threw the ball with a speed of 60 m/s vertically upwards.
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\( 2.30-\mathrm{kg} \) block starts from rest at the top of a \( 30.0^{\circ} \) incline and slides a distance of \( 1.70 \mathrm{~m} \) down the incline in \( 1.00 \mathrm{~s} \). (a) Find the magnit
(a). The magnitude of the acceleration of the block is approximately 4.9 m/s².
(b). The coefficient of kinetic friction between the block and the plane is approximately 0.304.
(c). The friction force acting on the block is approximately 10.40 N, and its direction is opposite to the direction of motion.
(d).The speed of the block after it has slid 2.30 m is approximately 6.03 m/s.
To find the answers to the given questions, we can use the principles of physics and the formulas for motion on an inclined plane.
(a) To find the magnitude of the acceleration of the block, we can use the formula for acceleration on an inclined plane:
a = g * sin(θ)
Where, g is the acceleration due to gravity and θ is the angle of the incline. In this case, θ = 30.0°. Using the value of g (9.8 m/s²), we can calculate the magnitude of the acceleration:
a = 9.8 m/s² * sin(30.0°)
a ≈ 4.9 m/s²
Therefore, the block accelerates at a rate of about 4.9 m/s2, which is a significant amount.
(b). To find the coefficient of kinetic friction between the block and the plane, we can use the formula:
μ = tan(θ) - a / g * cos(θ)
Where, μ is the coefficient of kinetic friction. In this case, θ = 30.0° and a = 4.9 m/s². Using these values, we can calculate the coefficient of kinetic friction:
μ = tan(30.0°) - 4.9 m/s² / 9.8 m/s² * cos(30.0°)
μ ≈ 0.304
Therefore, the block and the plane have a kinetic friction coefficient of about 0.304.
(c). To find the friction force acting on the block, we can use the formula: F_friction = μ * N
Where, F_friction is the friction force and N is the normal force. The normal force can be calculated using the formula:
N = m * g * cos(θ)
Where, m is the mass of the block and g is the acceleration due to gravity. In this case, m = 2.10 kg and θ = 30.0°. Using these values, we can calculate the normal force:
N = 2.10 kg * 9.8 m/s² * cos(30.0°)
N ≈ 34.22 N
Now we can calculate the friction force:
F_friction = 0.304 * 34.22 N
F_friction ≈ 10.40 N
Therefore, 10.40 N of friction force is exerted on the block, and it acts in the opposite direction to the motion.
(d) To find the speed of the block after it has slid 2.30 m, we can use the formula for the final velocity:
v_final = sqrt(2 * a * d)
Where, v_final is the final velocity, a is the acceleration, and d is the distance. Using the given values, we can calculate the final velocity: v_final = sqrt(2 * 4.9 m/s² * 2.30 m)
v_final ≈ 6.03 m/s
Therefore, after sliding 2.30 m, the block moves at a speed of roughly 6.03 m/s.
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Complete question is,
2.10−kg block starts from rest at the top of a 30.0∘ incline and slides a distance of 2.30 m down the incline in 1.00 s. (a) Find the magnitude of the acceleration of the block. m/s2 (b) Find the coefficient of kinetic friction between block and plane. (c) Find the friction force acting on the block. magnitude N direction (d) Find the speed of the block after it has slid 2.30 m. m/s
Please answer the following questions.
If the three masses used to balance out the ring are 300g, 120g, and 170g, what are the magnitudes of the three forces F1, F2, and F3 in Newtons?
We need to convert the masses to kilograms and use the equation F = mg. The magnitudes of the three forces F1, F2, and F3 are approximately 2.94 N, 1.18 N, and 1.67 N, respectively.
To determine the magnitudes of the three forces F1, F2, and F3, we need to convert the masses to kilograms and use the equation F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given the masses:
m1 = 300 g = 0.3 kg
m2 = 120 g = 0.12 kg
m3 = 170 g = 0.17 kg
Calculating the forces:
F1 = m1 * g = 0.3 kg * 9.8 m/s²
F2 = m2 * g = 0.12 kg * 9.8 m/s²
F3 = m3 * g = 0.17 kg * 9.8 m/s²
Evaluating these expressions:
F1 ≈ 2.94 N
F2 ≈ 1.18 N
F3 ≈ 1.67 N
Therefore, the magnitudes of the three forces F1, F2, and F3 are approximately 2.94 N, 1.18 N, and 1.67 N, respectively.
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A series RCL circuit is at resonance and contains a variable resistor that is set to 240Ω. The power dissipated in the circuit is 2.50 W. Assuming that the voltage remains constant, how much power is dissipated when the variable resistor is set to 470 Ω ? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.
When the variable resistor is set to 470 Ω, approximately 0.522 W of power is dissipated in the circuit. In a series RCL circuit at resonance, the power dissipated can be calculated using the formula.
P = I^2 * R
Where:
P is the power dissipated,
I is the current flowing through the circuit, and
R is the resistance.
Given that the power dissipated in the circuit is 2.50 W when the variable resistor is set to 240 Ω, we can write:
2.50 W = I^2 * 240 Ω
Solving for I:
I^2 = 2.50 W / 240 Ω
I^2 = 0.0104 A^2
I ≈ 0.102 A
Now, to find the power dissipated when the variable resistor is set to 470 Ω, we can use the same formula:
P = I^2 * R
Substituting the values:
P = (0.102 A)^2 * 470 Ω
P ≈ 0.522 W
Therefore, when the variable resistor is set to 470 Ω, approximately 0.522 W of power is dissipated in the circuit.
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Light with a vergence of −3.00D is incident on a lens in air. After leaving the lens, the vergence is −2.00D. The lens has a focal power of image will be formed.... the lens. −5.00D; a virtual; 20 centimeters in front of +1.00 D; a real; 50 centimeters behind +1.00 D; a virtual; 50 centimeters in front . +5.00 D; a real; 20 centimeters behind
A lens is a transparent object with a refractive index greater than that of air. When a light beam passes through the lens, it bends. A lens is also capable of forming an image by changing the direction of light. The image formed by a lens can be real or virtual, magnified or reduced, and upright or inverted.
The focal power of the lens is determined by its curvature and refractive index.The question states that light with a vergence of −3.00D is incident on a lens in air. After leaving the lens, the vergence is −2.00D. The lens has a focal power of 5D. Since the vergence of the light leaving the lens is negative, it means that the image formed is virtual.
The distance of the image from the lens is given by the formula:1/v - 1/u = 1/f
where v is the distance of the image from the lens, u is the distance of the object from the lens, and f is the focal length of the lens.
Solving for v gives:1/v = 1/f - 1/u1/v = 1/5 - (-1/3)1/v = 1/5 + 1/3v = 15/8 cm
Since the image is virtual, its distance from the lens is negative.
The image is located 15/8 cm in front of the lens. Hence, the correct option is +1.00 D; a virtual; 50 centimeters in front of the lens.
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In the same situation, what is true of the charge distribution on the outer surface of the spherical conductor? It is not uniform and not zero on average It is zero on average, but not uniform It is u
The charge distribution on the outer surface of the spherical conductor is zero on average but not uniform.
A spherical conductor of radius R is charged with total charge Q. The charge density is uniform throughout the volume of the conductor. Let us assume that the charge is uniformly distributed throughout the volume of the conductor with a charge density ρ. Now let us suppose that this charged conductor is made up of a material of negligible resistance, and that it is grounded or placed on a conductor of zero potential.
So, this is the situation that we are considering.
Therefore, the correct option is: It is zero on average, but not uniform.
Explanation: The electric field due to a charged sphere is equal to the charge contained within the sphere divided by the permittivity of free space (ϵo) multiplied by the radius of the sphere. That is,
E = Q/ 4πϵoR²
where Q is the total charge and R is the radius of the sphere.
Since the material is a conductor, the charge resides on the surface. This implies that the electric field just outside the conductor is perpendicular to the surface at every point, and that the electric field is zero inside the conductor.Since the electric field inside the conductor is zero, any charge that is inside the conductor must also be uniformly distributed on the surface of the conductor. The electric field just outside the surface of a uniformly charged sphere is given by
E = Q/ 4πϵoR²
It is zero on average, but not uniform. Therefore, we conclude that in the same situation, the charge distribution on the outer surface of the spherical conductor is zero on average but not uniform.
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A proton is released in a uniform electric field, and it experiences an electric force of 2.18×10
−14
N toward the south. What are the magnitude and direction of the electric field? 28. (II) Determine the magnitude and direction of the electric field at a point midway between a−8.0μC and a+5.8μC charge 8.0 cm apart. Assume no other charges are nearby. 33. (II) Calculate the electric field at one corner of a square 1.22 m on a side if the other three corners are occupied by 2.25×10
−6
C charges. 36. (II) Two point charges, Q
1
=−25μC and Q
2
=+45μC, are separated by a distance of 12 cm. The electric field at the point P (see Fig. 21−58 ) is zero. How far from Q
1
is P ?
According to the question (28.) Use E = (k * (|Q1| + |Q2|)) / (2 * r^2) for electric field at midpoint between charges. (33.) Sum contributions from each charge using E = (k * Q) / r^2 to find electric field at one corner of the square. (36.) Solve equation with E = (k * Q) / r^2 for distance from Q1 to point P where electric field is zero.
28. The magnitude and direction of the electric field at the midpoint between charges, -8.0 μC and +5.8 μC, with a distance of 0.08 m can be calculated as E = 5.6125 x 10^5 N/C towards the negative charge.
33. To find the electric field at one corner of a square with charges -8.0 μC and +5.8 μC, 8.0 cm apart (1.22 m sides), the net electric field is determined by calculating the electric field due to each charge individually and summing them up. The magnitude is approximately 1.045 x 10^5 N/C towards the negative charge.
36. To find the distance from charge Q1 to point P, where the electric field is zero due to charges -25 μC and +45 μC separated by 12 cm, we can use the principle of superposition and solve the equation to find that P is located approximately 6.67 cm from Q1.
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A spherical water drop, 1.20 μm in diameter, is suspended in calm air owing to a downward-directed atmospheric electric field E = 482 N/C.
What is the weight of the drop?
How many excess electrons does the drop have?
The acceleration due to gravity is approximately 9.8 m/s^2.
To find the weight of the water drop, we can use the formula:
weight = mass * acceleration due to gravity
First, let's calculate the mass of the water drop.
The volume of a spherical drop can be calculated using the formula:
volume = (4/3) * π * (radius)^3
Given that the diameter of the water drop is 1.20 μm, the radius can be calculated as half the diameter:
radius = 0.60 μm = 0.60 x 10^-6 m
Substituting the value of the radius into the volume formula:
volume = (4/3) * π * (0.60 x 10^-6 m)^3
Now, we can find the mass of the water drop using the formula:
mass = density * volume
The density of water is approximately 1000 kg/m^3.
Substituting the values into the formula:
mass = 1000 kg/m^3 * [(4/3) * π * (0.60 x 10^-6 m)^3]
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As you walk to class with a constant speed of 1.80 m/s, you are moving in a direction that is 22.9
∘
How much time does it take to change your displacement by 18.0 m east? north of east. Part B How much time does it take to change your displacement by 29.0 m north?
It takes approximately 16.11 seconds to change the displacement by 29.0 m north.To calculate the time it takes to change your displacement by 18.0 m east (north of east) and by 29.0 m north, we need to consider the horizontal and vertical components separately.
For the displacement of 18.0 m east (north of east), we can calculate the time using the horizontal component. The horizontal displacement is given by 18.0 m multiplied by the cosine of the angle (cos 22.9°).
Horizontal displacement = 18.0 m * cos(22.9°) = 16.130 m
To calculate the time, we can use the formula time = distance/speed:
Time = 16.130 m / 1.80 m/s ≈ 8.96 seconds
So, it takes approximately 8.96 seconds to change the displacement by 18.0 m east (north of east).
For the displacement of 29.0 m north, we can calculate the time using the vertical component. The vertical displacement is simply 29.0 m.
To calculate the time, we can again use the formula time = distance/speed:
Time = 29.0 m / 1.80 m/s ≈ 16.11 seconds
So, it takes approximately 16.11 seconds to change the displacement by 29.0 m north.
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sciencephysicsphysics questions and answersthe square surface shown in the figure measures 4.1 mm on each side. it is immersed in a uniform electric field with magnitude e= 2000 n/c and with field lines at an angle of 35∘ with a normal to the surface, as shown. take that normal to be "outward," as though the surface were one face of a box. calculate the electric flux through the surface. number
Question: The Square Surface Shown In The Figure Measures 4.1 Mm On Each Side. It Is Immersed In A Uniform Electric Field With Magnitude E= 2000 N/C And With Field Lines At An Angle Of 35∘ With A Normal To The Surface, As Shown. Take That Normal To Be "Outward," As Though The Surface Were One Face Of A Box. Calculate The Electric Flux Through The Surface. Number
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The square surface shown in the figure measures 4.1 mm on each side. It is immersed in a uniform electric field with magnitude E= 2000 N/C and with field lines at an angle of 35
∘
with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Number Units The electric field in a certain region of Earth's atmosphere is directed vertically down. At an altitude of 263 m the field has magnitude 87.6 N/C; at an altitude of 129 m, the magnitude is 153 N/C. Find the net amount of charge contained in a cube 134 m on edge, with horizontal faces at altitudes of 129 m and 263 m. Number Units
The electric flux through the surface is 2.58 × 10^-2 Nm²/C.
Side of square surface, a = 4.1 mm
Electric field, E = 2000 N/C
The angle between the field lines and the normal to the surface, θ = 35º
We can calculate the electric flux through the surface using the formula given below:
Φ = E . A . cos θ
Where A is the area of the surface.
The area of the square surface is given by,
A = a² = (4.1 × 10^-3 m)² = 1.68 × 10^-5 m²
Now, substitute the given values in the formula and solve for electric flux.
Φ = E . A . cos θΦ = (2000 N/C) (1.68 × 10^-5 m²) cos 35ºΦ = 2.58 × 10^-2 Nm²/C
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A fully charged 7.4 μFμF capacitor is connected in series with a 1.5×105 ΩΩ resistor.
What percentage of the original charge is left on the capacitor after 1.8 s of discharging? Express your answer as a percentage.
After 1.8 seconds of discharging, approximately 82.3% of the original charge remains on the capacitor.
To determine the percentage of the original charge remaining on the capacitor after 1.8 seconds of discharging, we need to calculate the charge on the capacitor at that time. The charge on a capacitor undergoing exponential discharge can be determined using the equation:
[tex]\(Q(t) = Q_0 \cdot e^{-\frac{t}{RC}}\)[/tex]
Where:
Q(t) is the charge on the capacitor at time t
Q₀ is the initial charge on the capacitor
t is the time
R is the resistance in the circuit
C is the capacitance
Initial charge (Q₀) = fully charged capacitor = 7.4 μF
Time (t) = 1.8 seconds
Resistance (R) = 1.5 × 10^5 Ω
Capacitance (C) = 7.4 μF
Plugging these values into the equation, we have:
[tex]\[Q(t) = 7.4 \mu F \cdot e^{-\frac{1.8 s}{1.5 \times 10^5 \Omega \cdot 7.4 \mu F}}\][/tex]
Simplifying this expression, we find:
[tex]\[Q(t) \approx 7.4 \mu F \cdot e^{-2.448 \times 10^{-11} s^{-1}}\][/tex]
Calculating the exponential term, we get:
Q(t) ≈ 7.4 μF * 0.823
Hence, the charge remaining on the capacitor after 1.8 seconds is approximately 6.084 μF. To find the percentage of the original charge remaining, we can divide this value by the initial charge (7.4 μF) and multiply by 100:
Percentage remaining = (6.084 μF / 7.4 μF) * 100 ≈ 82.3%
Therefore, approximately 82.3% of the original charge remains on the capacitor after 1.8 seconds of discharging.
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