(a) Orbital Speed V_A = √(G * M_e / (R_e + h_A)), (b) V_B = √(G * M_e / (R_e + h_B)).
(a) The orbital speed for satellite A can be calculated using the formula for the orbital speed of a satellite:
V_A = √(G * M_e / r_A)
where G is the gravitational constant, M_e is the mass of the Earth, and r_A is the radius of satellite A's orbit (which is the sum of the Earth's radius and the height of the orbit).
(b) The same formula can be used to calculate the orbital speed for satellite B, with r_B being the radius of satellite B's orbit.
The final expressions for the orbital speeds are:
(a) V_A = √(G * M_e / (R_e + h_A))
(b) V_B = √(G * M_e / (R_e + h_B))
where h_A and h_B are the heights of satellite A and satellite B above the Earth's surface, respectively.
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(8\%) Problem 7: Suppose you wanted to store \( 3 \mu \mathrm{C} \) of charge in a capacitor across a voltage of \( 120 \mathrm{~V} \). ( What capacitance is needed in \( \mathrm{nF} \) ? \[ C= \]
The value of the capacitance needed is determined as 25 nF.
What is the capacitance needed?If you wanted to store 3μC of charge in a capacitor across a voltage of 120 V, the value of the capacitance needed is calculated by applying the following formula.
C = Q/V
where;
Q is the chargeV is the voltage suppliedC is the capacitanceC = ( 3 x 10⁻⁶ C) / ( 120 V )
C = 2.5 x 10⁻⁸ F
C = 25 x 10⁻⁹ F
C = 25 nF
Thus, the value of the capacitance needed is determined as 25 nF.
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A 10 pole, three phase alternator has 60 slots. Each coil spans 5 slots. If the winding used is half-coil calculate the number of coils per phase.
For a 10-pole, three-phase alternator with 60 slots, each coil spanning 5 slots, and a half-coil winding, there are 12 coils per phase.
The number of coils per phase in a three-phase alternator can be calculated by considering the number of poles and the number of slots. In this case, we have a 10-pole alternator with 60 slots. Each coil spans 5 slots, and the winding used is half-coil.
To calculate the number of coils per phase, we can use the formula:
Number of coils per phase = (Number of slots) / (Number of slots spanned by each coil)
Given that each coil spans 5 slots, we can substitute this value into the formula:
Number of coils per phase = 60 / 5
Simplifying the equation:
Number of coils per phase = 12
Therefore, there are 12 coils per phase in this three-phase alternator.
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The impulse response of an LTI filter is given by h(t)=2e
−2t
u(t). (a) Determine the unit step response for this filter, that is find s(t) as the output of the filter when the input is u(t). (b) Determine the output, y(t), of the filter for an input x(t)=u(t+1)−u(t−3).
a. The unit step response s(t) for this filter is: s(t) = -e^(-2t) + 1, for t ≥ 0
b. The output y(t) of the filter for the input x(t) = u(t+1) - u(t-3) is:
y(t) = -e^(-2(t-1)) + 1 - (-e^(-2(t-3)) + 1), for t ≥ 0.
The impulse response of an LTI (Linear Time-Invariant) filter is given by h(t) = 2e^(-2t) u(t), where u(t) is the unit step function.
(a) To determine the unit step response for this filter, we need to convolve the impulse response h(t) with the unit step function u(t). The convolution operation is denoted by *, and it is defined as:
s(t) = h(t) * u(t)
In this case, h(t) = 2e^(-2t) u(t) and u(t) = u(t), so the convolution becomes:
s(t) = (2e^(-2t) u(t)) * u(t)
To perform the convolution, we need to integrate the product of h(t) and u(t) over the range from 0 to t:
s(t) = ∫[0,t] (2e^(-2τ) u(τ)) dτ
The unit step function u(τ) is 1 for τ >= 0 and 0 for τ < 0. Therefore, we can simplify the integral by considering two cases:
1. For 0 ≤ τ ≤ t:
s(t) = ∫[0,t] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t]
= -e^(-2t) + 1
2. For τ > t:
s(t) = ∫[0,t] (2e^(-2τ) u(τ)) dτ + ∫[t,∞] (2e^(-2τ) u(τ)) dτ
= ∫[0,t] (2e^(-2τ)) dτ + ∫[t,∞] 0 dτ
= -e^(-2τ) | [0,t] + 0
= -e^(-2t) + 1
Therefore, the unit step response s(t) for this filter is:
s(t) = -e^(-2t) + 1, for t ≥ 0
(b) To determine the output y(t) of the filter for the input x(t) = u(t+1) - u(t-3), we need to convolve the input signal x(t) with the impulse response h(t):
y(t) = x(t) * h(t)
Substituting the given values of x(t) and h(t) into the convolution equation, we have:
y(t) = (u(t+1) - u(t-3)) * (2e^(-2t) u(t))
Expanding the convolution and simplifying, we can split the integral into two parts:
y(t) = ∫[0,t] (2e^(-2τ) u(t+1-τ)) dτ - ∫[0,t] (2e^(-2τ) u(t-3-τ)) dτ
Considering two cases again:
1. For 0 ≤ τ ≤ t-1:
y(t) = ∫[0,t-1] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t-1]
= -e^(-2(t-1)) + 1
2. For 0 ≤ τ ≤ t-3:
y(t) = ∫[0,t-3] (2e^(-2τ)) dτ
= -e^(-2τ) | [0,t-3]
= -e^(-2(t-3)) + 1
Therefore, the output y(t) of the filter for the input x(t) = u(t+1) - u(t-3) is:
y(t) = -e^(-2(t-1)) + 1 - (-e^(-2(t-3)) + 1), for t ≥ 0.
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A real object in air is 50 cm away from a lens with a focal power of +5.00. D. What is the image vergence? −3.00D 0 −2.00 D 0 +3.00D 0 +7.00D
Vergence is the degree to which light rays are concentrated at the focal point, which is a physical quantity measured in diopters.
The image vergence is the vergence of light rays that are parallel to the axis of a lens that converge onto the lens and then leave it again. How do you determine the image vergence? The image vergence is determined by the formula:
V′ = V − D where V = the vergence of light incident on the lens and D = the power of the lens in diopters.
Since the object is real, it is located on the opposite side of the lens from the observer, and its image is formed on the same side as the observer. The distance between the lens and the real object is d = -50 cm since it is located on the opposite side of the lens.
The power of the lens in diopters is P = +5.00D. In this case, we have a positive power lens since it is a converging lens. Therefore, we need to use the formula:
V′ = V − D Where, V = the vergence of light incident on the lens and
D = the power of the lens in diopters V = 1/d V = 1/-50 cm V = -0.02 D
Now, we'll substitute the values in the equation: V′ = V − D⇒ V′ = -0.02 - 5⇒ V′ = -5.02D
The image vergence is -5.02 D. Answer: The correct option is -5.02 D.
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The electric field strength 7 cm from a very long charged wire is 1,945 N/C. What is the electric field strength 2 cm from the wire? Express your answer in N/C to the nearest 100 N/C.
The electric field strength 2 cm from the wire is approximately 158.78 N/C to the nearest 100 N/C.
For finding the electric field strength 2 cm from the wire, we can use the concept of inverse square law for electric fields. According to this law, the electric field strength is inversely proportional to the square of the distance from the charged wire.
Given that the electric field strength 7 cm from the wire is 1,945 N/C, can set up the following proportion:
[tex](7 cm)^2 : (2 cm)^2 = 1,945 N/C : x[/tex]
Simplifying the proportion,
49 : 4 = 1,945 N/C : x
Cross-multiplying and solving for x,
49 * x = 1,945 N/C * 4
x = (1,945 N/C * 4) / 49
x ≈ 158.78 N/C
Therefore, the electric field strength 2 cm from the wire is approximately 158.78 N/C to the nearest 100 N/C.
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hi i could use some help with the first part of this question A small \( 4 \mathrm{~kg} \) block is accelerated from rest on a flat surface by a compressed spring \( (k=636 \mathrm{~N} / \mathrm{m}) \) along a frictionless, horizontal surface. The block leaves t
The small 4 kg block is accelerated from rest on a flat surface by a compressed spring .
When a spring is compressed and then released, it exerts a force known as the spring force. This force can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
In this scenario, the spring constant is given as 636 N/m. To determine the force exerted by the compressed spring, we need to know the displacement of the spring. Unfortunately, the displacement value is not provided in the question. Once the displacement is known, we can calculate the force using the formula F = k * x, where F is the force, k is the spring constant, and x is the displacement.
The force exerted by the spring is responsible for accelerating the 4 kg block. According to Newton's second law of motion, the acceleration of an object is equal to the net force acting on it divided by its mass. Therefore, the force exerted by the spring divided by the mass of the block will give us the acceleration of the block.
Please provide the displacement value of the spring so that we can calculate the force and subsequently the acceleration of the block.
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According to Equation 20.7, an ac voltage V is given as a function of time t by V=V
o
sin2πft, where V
0
is the peak voltage and f is the frequency (in hertz). For a frequency of 47.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peakvalue?
The smallest value of time at which the voltage equals one-half of the peak value, for a frequency of 47.7 Hz, is approximately 0.0105 seconds.
According to Equation 20.7, the voltage V as a function of time t is given by V = V0 sin(2πft), where V0 is the peak voltage and f is the frequency. We want to find the smallest value of time at which the voltage equals one-half of the peak value. In other words, we need to solve the equation V = (1/2)V0 for t.
Substituting the given frequency f = 47.7 Hz into the equation, we have:
(1/2)V0 = V0 sin(2π(47.7)t)
Dividing both sides of the equation by V0, we get:
1/2 = sin(2π(47.7)t)
To find the smallest value of time at which the equation is satisfied, we can take the inverse sine (sin^(-1)) of both sides:
sin^(-1)(1/2) = 2π(47.7)t
Simplifying further, we have:
t = sin^(-1)(1/2) / (2π(47.7))
Using a calculator, we can evaluate sin^(-1)(1/2) to be approximately 30 degrees or π/6 radians.
Plugging in this value, we get:
t ≈ (π/6) / (2π(47.7))
Simplifying, we find:
t ≈ 1 / (2(47.7))
t ≈ 0.0105 seconds
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"Three forces are applied to an object with a mass of 2 kg. These
are the only forces acting on the object. F1 has a magnitude of 6N
and points directly to the left. F2 has a magnitude of 5N and
points"
The net force acting on the object is 1 N to the left.
Three forces are applied to an object with a mass of 2 kg. These are the only forces acting on the object. F1 has a magnitude of 6N and points directly to the left. F2 has a magnitude of 5N and points directly upwards. F3 has a magnitude of 4N and points directly downwards.
The first step is to resolve the forces in the horizontal direction since F1 acts horizontally.
Using Pythagoras theorem, we can find that the horizontal component of F2 and the horizontal component of F3 are equal to 3 N. Therefore, the total horizontal force acting on the object is 3N to the left (6N-3N=3N).
Hence, the net force acting on the object is 1 N to the left (3N-2N=1N).
Therefore, the object will move towards the left direction with an acceleration of 0.5 m/s² which is determined using the formula F=ma, where F is the net force acting on the object, m is the mass of the object and a is the acceleration of the object.
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A cement block accidentally falls from rest from the ledge of a 84.5-m-high building. When the block is 19.1 m above the ground, a man, 1.70 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
To determine the maximum time the man has to get out of the way of a falling cement block, we can calculate the time it takes for the block to fall from a height of 19.1 m to the ground.
Using the equations of motion, we can find the time by considering the vertical distance traveled by the block. The correct answer depends on the acceleration due to gravity and the initial height of the block.
The vertical distance traveled by the block is the difference between the initial height (84.5 m) and the final height (19.1 m). Using the equation of motion,
h = ut + (1/2)gt², where
h is the vertical distance,
u is the initial velocity (0 m/s in this case),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
t is the time,
we can calculate the time it takes for the block to fall.
The equation becomes:
19.1 = 0 + (1/2)(9.8)t²
Simplifying the equation:
9.8t² = 19.1 × 2
t² = (19.1 × 2) / 9.8
t² ≈ 3.898
t ≈ √3.898
t ≈ 1.97 seconds
Therefore, the maximum time the man has to get out of the way is approximately 1.97 seconds. During this time, the block will fall from a height of 19.1 m to the ground. It's crucial for the man to move quickly to avoid the falling block and ensure his safety.
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In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of 9.06 m/s in 2.31 s. Assuming that the player accelerates uniformly, determine the distance he runs.
To determine the distance the basketball player runs, we can use the equation of motion. The basketball player runs approximately 5.41 meters.
To determine the distance the basketball player runs, we can use the equation of motion:
[tex]s = ut +\frac{1}{2} at^{2}[/tex]
Where:
s = distance
u = initial velocity (0 m/s, as the player starts from rest)
a = acceleration
t = time is taken (2.31 s)
Since the player starts from rest, the initial velocity (u) is 0 m/s. We need to find the acceleration (a) to calculate the distance.
Using the equation of motion:
v = u + at
9.06 = 0 + a x 2.31
Simplifying the equation:
9.06 = 2.31a
a = 9.06/2.31
a = 3.925 m/s^2
Now, we can substitute the values of u, a, and t into the distance equation:
s = 0 x 2.31 + 1/2 x 3.925 x (2.31)^2
s = 5.41 m
Therefore, the basketball player runs approximately 5.41 meters.
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A rocket undergoes a constant acceleration of 2.6 m/s
2
starting from rest. What is the distance traveled, in meters, in 3.4 minutes? (Round off your answer to the ones.)
The distance traveled by the rocket in 3.4 minutes is 54,091.2 m
Given :
Acceleration of the rocket is 2.6 m/s².
Time for which the rocket moves is 3.4 minutes or 204 seconds (1 minute = 60 seconds).
We need to find the distance traveled by the rocket.
We can use the following kinematic equation :
distance = initial velocity × time + 0.5 × acceleration × time²
As the rocket starts from rest, initial velocity (u) is zero.
Therefore, distance = 0 + 0.5 × 2.6 × (204)²
distance = 0 + 0.5 × 2.6 × 41,616
distance = 0 + 54,091.2
Therefore, the distance traveled by the rocket is 54,091.2 m (rounding off to the nearest meter).
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A high jumper of mass 70.1 kg consumes a meal of 4.20 × 10^3 kcal prior to a jump. If 3.30% of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump?
The athlete could jump approximately 827.9 meters high with the gravitational potential energy obtained from consuming the meal.
To determine how high the athlete could jump, we need to calculate the gravitational potential energy (GPE) that can be obtained from the consumed meal and then convert it to the height.
First, let's convert the energy consumed from kilocalories (kcal) to joules (J):
1 kcal = 4184 J
Energy consumed = [tex]4.20 * 10^3[/tex] kcal * 4184 J/kcal
Energy consumed = [tex]1.75 * 10^7[/tex] J
Next, we need to find the gravitational potential energy (GPE) that can be obtained from the consumed energy. We know that 3.30% of the energy can be converted to GPE:
GPE = 0.0330 × Energy consumed
GPE = [tex]0.0330 * 1.75 * 10^7[/tex] J
GPE = [tex]5.775 * 10^5[/tex] J
To convert the GPE into height, we can use the formula:
GPE = mgh
Where:
m is the mass of the jumper (70.1 kg),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
h is the height.
Rearranging the formula, we can solve for h:
h = GPE / (mg)
h = (5.775 * 10⁵ J) / (70.1 kg * 9.8 m/s²)
Calculating the height:
h ≈ 827.9 meters.
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Light of wavelength 630 nm falls on two slits and produces an Part A interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away. What is the separation of the two slits? Express your answer using two significant figures. X Incorrect; Try Again; 4 attempts remaining
The separation of the two slits is 0.0158 mm (approx).
Given,Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright fringe is 40 mm from the central fringe on a screen 3.0 m away.Concept used,The formula for calculating the separation between the two slits is given by: dsinθ = m λ where, m = 1,2,3,...θ = angle of diffractiond = separation between the slitsλ = wavelength of the lightThe path difference between the waves of light emerging from two slits is given by: d sinθ = mλ where, d = separation between the slitsλ = wavelength of lightθ = angle of diffraction.The third order fringe means m = 3. Hence, d sinθ = 3λHere, λ = 630 nm = 6.3 × 10⁻⁷ m. Therefore,3d sinθ = 6.3 × 10⁻⁷ ...................(1)
We know that angle of diffraction, θ can be given by, θ = tan⁻¹(y/L) where y is the fringe width and L is the distance between the screen and the slits. Here, y = 40 mm = 0.04 mL = 3.0 m Therefore, θ = tan⁻¹ (0.04/3)Now, substitute this value of θ in equation (1), we get:3d × (0.04/3) = 6.3 × 10⁻⁷Or, d = (6.3 × 10⁻⁷ )/0.04The value of d is, d = 1.58 × 10⁻⁵ m or 0.0158 mm (approx).Hence, the separation of the two slits is 0.0158 mm (approx).
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A periodic sawtooth train with fundamental period T
0
, amplitude A, and pulse duration τ
0
, over the fundamental period is expressed as x
1
(t)={
τ
A
(t+
2
τ
)
0
−
2
τ
≤t<
2
τ
,
otherwise
τ
0
Based on the analysis equation, derive the coefficients of the complex exponential Fourier series of the sawtooth train. (Note: The answer for the FS coefficients can be found in the Table 3.2 of the lecture notes.) (15 marks) (b) Consider the periodic signal x
2
(t) with fundamental period T
0
, whose expression over the period is given as x
2
(t)=
⎩
⎨
⎧
τ
2A
(t+
2
τ
)
−
τ
2A
(t−
2
τ
)
0
−
2
τ
≤t<0
0≤t<
2
τ
otherwise
Express x
2
(t) in terms of x
1
(t) and then use the properties of the Fourier series, or otherwise, to find the Fourier series coefficients of the periodic signal x
2
(t). (10 marks) (c) Find the Fourier series of the periodic signal x
3
(t) with period 6 . x
3
(t)=
⎩
⎨
⎧
2(t+2)
2
−2(t−2)
0
−2≤t<−1
−1≤t<1
1≤t<2
otherwise
To derive the coefficients of the complex exponential Fourier series of the sawtooth train x1(t), we need to express it as a sum of complex exponential functions.
First, let's determine the period of x1(t) from the given information.
The fundamental period T0 of the sawtooth train is the duration between two consecutive pulse starts, which is 2τ.
The Fourier series representation of x1(t) can be written as:
[tex]x1(t) = Σ Cn * e^(jnω0t)[/tex]
where Cn are the Fourier series coefficients, [tex]ω0 = 2π/T0[/tex]is the fundamental angular frequency, and [tex]j = √(-1)[/tex] is the imaginary unit.
To find the coefficients Cn, we need to calculate the integral of x1(t) multiplied by the complex conjugate of [tex]e^(jnω0t)[/tex]over a period T0.
Let's compute this integral step by step for the given expression of x1(t):
[tex]∫[τA(t+2τ)] * e^(-jnω0t) dt, -2τ ≤ t < 2τ[/tex]
We can simplify the expression by shifting the variable of integration, let's define [tex]u = t + 2τ[/tex]:
[tex]∫[τA(u)] * e^(-jnω0(u-2τ)) du, 0 ≤ u < 4τ[/tex]
Now, we can expand the exponential term using Euler's formula:
[tex]∫[τA(u)] * [cos(nω0u) - jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Since the interval of integration is one period T0, the cosine term integrates to zero:
[tex]∫[τA(u)] * [-jsin(nω0u)] * e^(jn2π) du, 0 ≤ u < 4τ[/tex]
Next, we can distribute the j and the e^(jn2π) term:
[tex]-j * e^(jn2π) * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
The term e^(jn2π) is equal to 1, so it can be omitted:
[tex]-j * ∫[τA(u) * sin(nω0u)] du, 0 ≤ u < 4τ[/tex]
Finally, we substitute back t for u:
[tex]-j * ∫[τA(t+2τ) * sin(nω0(t+2τ))] dt, -2τ ≤ t < 2τ[/tex]
The integral of the product of a sine function and a periodic function over one period is zero, so the integral evaluates to zero.
Therefore, for[tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient Cn is zero.
For all other values of t, the value of x1(t) is τ0.
Hence, for [tex]-2τ ≤ t < 2τ[/tex], the Fourier series coefficient C0 is τ0.
In summary, the coefficients of the complex exponential Fourier series of the sawtooth train x1(t) are:
[tex]C0 = τ0 (for -2τ ≤ t < 2τ)[/tex]
[tex]Cn = 0 (for all n ≠ 0)[/tex]
The actual coefficients may vary depending on the specific values of τ0 and A.
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Suppose that the radius of a disk R=25 cm, and the total charge distributed uniformly all over the disk is Q=8.0×10−6C. Use the exact result to calculate the electric field 1 mm from the center of the disk. N/C Use the exact result to calculate the electric field 3 mm from the center of the disk. N/C Does the field decrease significantly? Yes No
Given that radius of the disk, R = 25 cmTotal charge distributed uniformly all over the disk, Q = 8.0 × 10⁻⁶ C We need to calculate the electric field 1 mm from the center of the disk and 3 mm from the center of the disk.
The formula to calculate the electric field due to a disk is, E = σ/2ε₀ [1 - (z/√(z² + R²))]Where, σ is the surface charge density, ε₀ is the permittivity of free space, and z is the perpendicular distance from the center of the disk. The surface charge density, σ = Q/πR² = (8 × 10⁻⁶ C)/(π × (25 × 10⁻² m)²) = 2.03 × 10⁻⁷ C/m²Electric field 1 mm from the center of the disk, z = 1 mm = 0.001 m E₁ = (2.03 × 10⁻⁷)/(2 × 8.85 × 10⁻¹²) [1 - (0.001/√(0.001² + 0.25²))] = 6.52 × 10⁴ N/C Electric field 3 mm from the center of the disk, z = 3 mm = 0.003 m E₂ = (2.03 × 10⁻⁷)/(2 × 8.85 × 10⁻¹²) [1 - (0.003/√(0.003² + 0.25²))] = 2.33 × 10⁴ N/C Electric field decreases from 6.52 × 10⁴ N/C to 2.33 × 10⁴ N/C when the distance increases from 1 mm to 3 mm. Therefore, the field decreases significantly.
When the radius of a disk R=25 cm, and the total charge distributed uniformly all over the disk is Q=8.0×10−6C, the electric field at a distance of 1 mm from the center of the disk is 6.52 × 10⁴ N/C and the electric field at a distance of 3 mm from the center of the disk is 2.33 × 10⁴ N/C. The electric field decreases significantly when the distance increases from 1 mm to 3 mm. Therefore, the field decreases significantly.
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 39.3rad/s. Find the angular displacement Δθ of the tub during a spin of 91.3 s, expressed both in radians and in revolutions. Δθ= Δθ=
The angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
To find the angular displacement Δθ of the tub during a spin of 91.3 s, we can use the formula:
Δθ = ωt
where Δθ is the angular displacement in radians, ω is the angular velocity in rad/s, and t is the time in seconds.
Given:
ω = 39.3 rad/s
t = 91.3 s
Substituting the values into the formula, we have:
Δθ = 39.3 rad/s * 91.3 s
Calculating the product:
Δθ ≈ 3589.09 rad
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians.
To convert this angular displacement to revolutions, we can use the conversion factor: 1 revolution = 2π radians.
Δθ in revolutions = Δθ in radians / (2π)
Δθ in revolutions = 3589.09 rad / (2π)
Calculating:
Δθ ≈ 570.46 revolutions
Therefore, the angular displacement of the tub during a spin of 91.3 s is approximately 3589.09 radians or 570.46 revolutions.
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You and your lab partner, buoyed by the success of your first rock-dropping experiment, make a new plan. This time your lab partner goes to the fourth floor balcony of the MTSC building, which is 11.9 meters above the ground, while you wait at the ground below. Your partner throws a rock downward with an initial speed of 7.70 m/s. Find: (a) the speed of the rock as it hits the ground; (b) the time the rock is in freefall.
The speed of the rock as it hits the ground can be determined using the equations of motion. Since the rock is thrown downward, its initial velocity is negative.
The acceleration due to gravity is constant (taking downward direction as negative). The final velocity of the rock when it hits the ground is 0 m/s since it comes to a stop. We can use the equation [tex]v = v_0 + at[/tex], where v is the final velocity, [tex]v_0[/tex] is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we have:
0 = -7.70 m/s + ([tex]-9.8 m/s^2[/tex])t
Solving for t, we find:
t = 7.70 m/s / [tex]9.8 m/s^2[/tex] ≈ 0.7857 s
The time it takes for the rock to hit the ground is approximately 0.7857 seconds.
To find the speed of the rock as it hits the ground, we can use the equation v = v0 + at. Plugging in the values, we have:
v = -7.70 m/s + ([tex]-9.8 m/s^2[/tex])(0.7857 s)
v ≈ -14.54 m/s
The negative sign indicates that the rock has a downward velocity. Taking the absolute value, the speed of the rock as it hits the ground is approximately 14.54 m/s.
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A baseball player hits a fly ball that has an initial velocity for which the horizontal component is 30 m/s and the vertical component is 40 m/s. What is the speed of the ball at the highest point of its flight?
1. 50m/s
2. Zero
3. 30m/s
4. 40 m/s
The speed of the ball at the highest point is equal to the magnitude of its horizontal component of velocity, which is 30 m/s.
At the highest point of its flight, the vertical component of the ball's velocity becomes zero while the horizontal component remains unchanged. The speed of the ball at the highest point can be found by calculating the magnitude of the velocity vector.
Using the Pythagorean theorem, we can calculate the magnitude of the velocity vector:
speed = √((horizontal component)^2 + (vertical component)^2)
speed = √((30 m/s)^2 + (0 m/s)^2)
speed = √(900 m^2/s^2)
speed = 30 m/s.
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A ball thrown straight up with an initial velocity of +12 m/s. Find its position, velocity, and acceleration at (A) 1.0 s, (B) the maximum height, (C) 2.0 s, and (D) the moment right before it is caught at the same height it was thrown from. d=v
u
t+0.5gt
2
Vf=v
a
+gt where g=−10 m/s/s 13. For the ball in the previous problem, how much time does it take to reach the maximum height? 14. Make a table of the velocities of an object at the end of each second for the first 5 s of free-fall from rest. Assume in this problem that down is the positive direction. Therefore your velocities will all be positive. a. Use the data in your table to plot a velocity time graph below in the grid shown. b. What does the total area under the curve represent? 12. A ball thrown straight up with an initial velocity of +12 m/s. Find its position, velocity, and acceleration at (A) 1.0 s, (B) the maximum height, (C) 2.0 s, and (D) the moment right before it is caught at the same height it was thrown from. d=v
u
t+0.5gt
2
Vf=v
a
+gt where g=−10 m/s/s 13. For the ball in the previous problem, how much time does it take to reach the maximum height? 14. Make a table of the velocities of an object at the end of each second for the first 5 s of free-fall from rest. Assume in this problem that down is the positive direction. Therefore your velocities will all be positive. a. Use the data in your table to plot a velocity time graph below in the grid shown. b. What does the total area under the curve represent?
A ball is thrown straight upwards with an initial velocity of +12 m/s. The velocity, position, and acceleration at different times can be calculated as shown below: At time, t = 1 s, v = u + at= 12 - 10(1) = 2 m/s
A ball is thrown straight upwards with an initial velocity of +12 m/s. The velocity, position, and acceleration at different times can be calculated as shown below: At time, t = 1 s, v = u + at= 12 - 10(1) = 2 m/s
The final velocity,[tex]v_f = v_i[/tex] + at= 12 - 10(1) = 2 m/s
The displacement, s = [tex]v_i[/tex]* t + (1/2) * a * t²= 12(1) + (1/2)(-10)(1)²= 7 m
At the maximum height, v = 0. Therefore, t = [tex]v_f[/tex]/g= 2/-10= 0.2 s
The displacement, s = [tex]v_i[/tex]* t + (1/2) * a * t²= 12(0.2) + (1/2)(-10)(0.2)²= 1.2 m
At time, t = 2 s, v = [tex]v_i[/tex]+ at= 12 - 10(2) = -8 m/s
The final velocity, [tex]v_f = v_i[/tex] + at= 12 - 10(2) = -8 m/s
The displacement, s = [tex]v_i[/tex] * t + (1/2) * a * t²= 12(2) + (1/2)(-10)(2)²= 2 m
At the moment right before it is caught at the same height, v = 0. Therefore, t = [tex]v_f[/tex] /g= -12/-10= 1.2 s
The displacement, s = [tex]v_i[/tex] * t + (1/2) * a * t²= 12(1.2) + (1/2)(-10)(1.2)²= 7.2 m
The velocity of the object at the end of each second for the first 5 s of free-fall from rest can be calculated as shown below: Time, t (s)Velocity, v (m/s)10+0=001+(-10)=9-19+(-10)=8-27+(-10)=7-35+(-10)=6
a) The velocity-time graph is shown below: b) The total area under the curve represents the displacement of the object.
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Find the Normal force and the acceleration experienced by a block of 6 kg being pulled by a force of 25 N at an angle of 30
∘
with the floor
The normal force experienced by the block is approximately 46.3 N, and the acceleration of the block is approximately 3.61 m/s².
To find the normal force and the acceleration experienced by the block, we need to consider the forces acting on the block. Let's break down the forces involved:
Force of gravity (weight):
The force of gravity acting on the block can be calculated using the formula: weight = mass * gravity.
Given the mass of the block is 6 kg and the acceleration due to gravity is approximately 9.8 m/s², the weight of the block is: weight = 6 kg * 9.8 m/s² = 58.8 N.
Vertical component of the applied force:
The applied force is at an angle of 30 degrees with the floor. We need to find the vertical component of the applied force, which contributes to the normal force. The vertical component can be calculated as: vertical force = applied force * sin(angle).
Given the applied force is 25 N and the angle is 30 degrees, the vertical component of the applied force is: vertical force = 25 N * sin(30°).
Normal force:
The normal force is the perpendicular force exerted by the floor on the block, which counteracts the vertical force due to the applied force. The normal force can be calculated as: normal force = weight - vertical force.
Horizontal component of the applied force:
The applied force also has a horizontal component, which contributes to the acceleration of the block. The horizontal component can be calculated as: horizontal force = applied force * cos(angle).
Given the applied force is 25 N and the angle is 30 degrees, the horizontal component of the applied force is: horizontal force = 25 N * cos(30°).
Frictional force:
If there is no mention of friction, we can assume a frictionless scenario, and therefore, there is no frictional force.
Acceleration:
Using Newton's second law of motion, we can relate the net force acting on the block to its acceleration: net force = mass * acceleration.
The net force can be calculated as: net force = horizontal force.
Given the mass of the block is 6 kg, we have: horizontal force = 6 kg * acceleration.
Now, let's calculate the values:
Calculating the vertical component of the applied force:
vertical force = 25 N * sin(30°) ≈ 12.5 N
Calculating the normal force:
normal force = weight - vertical force
normal force = 58.8 N - 12.5 N ≈ 46.3 N
Calculating the horizontal component of the applied force:
horizontal force = 25 N * cos(30°) ≈ 21.65 N
Calculating the acceleration:
horizontal force = 6 kg * acceleration
21.65 N = 6 kg * acceleration
acceleration = 21.65 N / 6 kg ≈ 3.61 m/s²
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A heavy block, labeled " A ", is sitting on a table. On top of that block is a lighter block, labeled "B" as shown in the figure at the right. For the first parts of this problem you are asked to identify the direction of forces in this system under various circumstances. The labels in the subscripts indicate: A= block A,B= block B,F= finger, T= table. Specify the direction in your answers using the following notation: - R means points to the right - Lmeans points to the left - U means point up - D means points down - O indicates there is no such force at the instant specified 1. You start pushing on block A as shown, but it is too heavy and does not move. While you are pushing on block A but while it is not moving, specify the direction of the following normal ( N ) and frictional (f) forces between the various objects indicated. (a) N
A→B
(b) f
Y→A
(c) f
A→B
(d) N
F→A
(e) f
t→A
(f) N
0→A
2. Now you push a little harder and the block begins to move. Block B moves with it without slipping. While the blocks are speeding up, specify the direction of the following forces between the various objects indicated. (a) N
i→h
(b) f
T→A
(c) f
A→B
(d) N
f→A
(e) f
t→A
(f) N
0→A
3. Now you push so that the blocks move at a constant velocity. Block B moves with A without slipping. While the blocks are moving at a constant speed specify the direction of the following forces between the various objects indicated. (a) N
A→B
(b) f
T→A
(c) f
A→B
(d) N
f→A
(e) f
B−A
(f) N
B→A
In the given system of blocks (A and B) placed on a table, the directions of normal and frictional forces are determined for both motion and rest situations below:
1. (a) N A→B - Normal force will be pointing down from block A to block B, which is the reaction force to the weight of block B exerted on A. (b) f Y→A - Frictional force will be pointing to the left and it is between the surface of the table and blocks A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N F→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f t→A - Frictional force will be pointing to the right and it is between the surface of the table and block A, which opposes the direction of motion. (f) N 0→A - No force acting in the upward direction on block A.2. (a) N i→h - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (b) f T→A - Frictional force will be pointing to the left and it is between the surface of the table and block A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N f→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f t→A - Frictional force will be pointing to the left and it is between the surface of the table and block A, which opposes the direction of motion. (f) N 0→A - No force acting in the upward direction on block A.3. (a) N A→B - Normal force will be pointing down from block A to block B, which is the reaction force to the weight of block B exerted on A. (b) f T→A - Frictional force will be pointing to the left and it is between the surface of the table and blocks A, which opposes the direction of motion. (c) f A→B - Frictional force will be pointing to the right and it is between the surface of block A and block B, which opposes the direction of motion. (d) N f→A - Normal force will be pointing up from the table to block A, which is the reaction force to the weight of block A exerted on the table. (e) f B−A - Frictional force will be pointing to the left and it is between the surface of block B and block A, which opposes the direction of motion. (f) N B→A - Normal force will be pointing up from block B to block A, which is the reaction force to the weight of block B exerted on block A. Thus, the directions of normal and frictional forces in the given system of blocks (A and B) placed on a table are identified while in motion and at rest.For more questions on frictional forces
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Find the time for one complete vibration.
B.) Find the force constant of the spring.
C.) Find the maximum speed of the mass.
D.) Find the maximum magnitude of force on the mass.
E.) Find the position of the mass at t=1.00s;
F.) Find the speed of the mass at t=1.00s;
G.) Find the magnitude of acceleration of the mass at t=1.00s;
H.) Find the magnitude of force on the mass at t=1.00s;
To find the time for one complete vibration, force constant of the spring, maximum speed of the mass, maximum magnitude of force on the mass, position of the mass at t=1.00s, speed of the mass at t=1.00s, magnitude of acceleration of the mass at t=1.00s, and magnitude of force on the mass at t=1.00s, we need more information about the system you are referring to.
The time for one complete vibration, also known as the period (T), can be found using the formula T = 2π√(m/k), where m is the mass of the object attached to the spring and k is the force constant of the spring.
The force constant of the spring (k) can be calculated by dividing the force applied to the spring (F) by the displacement caused by the force (x). Therefore, k = F/x.
The maximum speed of the mass can be determined using the equation v = ωA, where ω is the angular frequency of the oscillation and A is the amplitude of the oscillation.
The maximum magnitude of force on the mass can be found using the formula Fmax = kA, where A is the amplitude of the oscillation.
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The thin plastic rod shown in the figure has length L=11.0 cm and a nonuniform linear charge density λ=cx,wherec=40.6pC/m
2
. Nith V=0 at infinity, find the electric potential at point P
1
on the axis, at distance d=4.10 cm from one end.
The electric potential at point P₁, located at a distance of 4.10 cm from one end of the rod, can be determined by integrating the contributions from all infinitesimally small elements of the rod.
To find the electric potential at point P₁ on the axis, we can use the principle of superposition. We need to consider the contribution to the potential from each infinitesimally small element of the rod and integrate over the entire length.
The electric potential due to an infinitesimally small element of length dx at a distance x from P₁ is given by dV = k * λ * dx / r, where k is the electrostatic constant and r is the distance from the element to P₁.
The linear charge density λ = cx, where c = 40.6 pC/m². Therefore, λ = 0.406x nC/m².
The distance from the element to P₁ is r = sqrt(x² + d²), where d = 4.10 cm = 0.041 m.
The electric potential at P₁ is obtained by integrating the contributions from all the elements:
V = ∫(k * λ * dx / r) from x = 0 to x = L.
V = ∫(k * 0.406x * dx / sqrt(x² + d²)) from x = 0 to x = L.
Solving this integral will give us the electric potential at point P₁.
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Water is poured into a container that has a leak, The mass m of the water is glven as a function of time t by m=6.00t−0.0−3.30t+17.00, with t2−0. m in grams, and t in seconds (a) AL what the is the water mass greatest? (b) What is that greatest mass? a (e) In kiloarams per minute, what is the rate of mais change at t=2.00 s? kgimin (d) In kiloarams per minute, what is the rate of mass change at t=5.00 s? karmin
Given mass of water in the container, m = 6.00t - 0.0 - 3.30t + 17.00 and t^2 - 0. The mass of the water m is given in grams and time t is in seconds.(a) For the greatest mass of water in the container, we need to differentiate the given mass expression with respect to t and equate it to zero.
Let's do it as follows:dm/dt = 6.00 - 6.60t = 0=> 6.60t = 6.00=> t = 6.00 / 6.60 = 0.909 sec Therefore, the water mass is maximum at 0.909 s.(b) For maximum mass, we need to put t = 0.909 s in the given mass expression, we getm = 6.00t - 0.0 - 3.30t + 17.00=> m = 6.00 (0.909) - 3.30 (0.909)^2 + 17.00=> m = 5.454 - 2.831 + 17.00=> m = 19.62 g Therefore, the maximum mass is 19.62 g.(c) In kilograms per minute, what is the rate of mass change at t = 2.00 s?To find the rate of mass change, we need to differentiate the given mass expression with respect to t and find the value of dm/dt at t = 2.00 s. Let's do it as follows:dm/dt = 6.00 - 6.60tAt t = 2.00 s,dm/dt = 6.00 - 6.60 (2.00) = -6.60 g/s The rate of mass change at t = 2.00 s is -6.60 g/s.Now, to convert it into kg/min, we will follow these steps:- 6.60 g/s = -6.60 x 60 x 0.001 kg/min (multiply by -1 to get the value in positive)- 6.60 g/s = -0.396 kg/min
Therefore, the rate of mass change at t = 2.00 s is 0.396 kg/min.(d) In kilograms per minute, what is the rate of mass change at t = 5.00 s?At t = 5.00 s,dm/dt = 6.00 - 6.60 (5.00) = -27.60 g/s The rate of mass change at t = 5.00 s is -27.60 g/s.Now, to convert it into kg/min, we will follow these steps:- 27.60 g/s = -27.60 x 60 x 0.001 kg/min (multiply by -1 to get the value in positive)- 27.60 g/s = -1.656 kg/min Therefore, the rate of mass change at t = 5.00 s is 1.656 kg/min. (Note: The rate of mass change is negative at both t = 2.00 s and t = 5.00 s because the water is leaking out of the container.)Hence, the long answer to the given problem is as follows:(a) The water mass is maximum at 0.909 s.(b) The maximum mass is 19.62 g.(c) The rate of mass change at t = 2.00 s is 0.396 kg/min.(d) The rate of mass change at t = 5.00 s is 1.656 kg/min.
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A rocket has 13653 N of propulsion and experience a constant kinetic friction of 9206 N. The rocket accelerates at a rate of 14 m/s/s. What is the mass of the rocket in kg?
To find the mass of the rocket, we can use Newton's second law of motion, the mass of the rocket is approximately 317.64 kg.
To find the mass of the rocket, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
Given:
Propulsion force (F_propulsion) = 13653 N
Kinetic friction force (F_friction) = 9206 N
Acceleration (a) = 14 m/s²
The net force acting on the rocket can be calculated by subtracting the kinetic friction force from the propulsion force:
Net force (F_net) = F_propulsion - F_friction
Substituting the given values:
F_net = 13653 N - 9206 N
= 4447 N
Now, we can use Newton's second law to find the mass (m):
F_net = m * a
4447 N = m * 14 m/s²
Dividing both sides of the equation by 14 m/s²:
m = 4447 N / 14 m/s²
m ≈ 317.64 kg
Therefore, the mass of the rocket is approximately 317.64 kg.
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The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.940 s until the car engine starts. How much charge passes through the starter motor? Express your answer with the appropriate units. Part B How far does an electron travel along the wire while the starter motor is on? Express your answer with the appropriate units. X Incorrect; Try Again; 3 attempts remaining
The charge passing through the starter motor is 32.4 C (coulombs), and an electron travels approximately 0.59 cm (centimeters) along the wire during the operation of the starter motor.
To calculate the charge passing through the starter motor, we can use the formula Q = I * t, where Q represents the charge, I is the current, and t is the time. In this case, the current drawn by the starter motor is 180 A, and it runs for 0.940 s. Plugging these values into the formula, we get Q = 180 A * 0.940 s = 169.2 C. Therefore, approximately 169.2 C or 32.4 C of charge passes through the starter motor.
To find the distance an electron travels along the wire, we need to calculate the length of the wire. The wire's diameter is given as 4.60 mm, and we can use the formula for the circumference of a circle, C = π * d, where C is the circumference and d is the diameter. Substituting the given value, we find C = π * 4.60 mm = 14.45 mm. Converting mm to cm, we get C ≈ 1.445 cm. Since the electron travels along the wire's length, which is 1.2 m or 120 cm, the distance the electron travels is approximately 1.445 cm * (120 cm / 1.445 cm) = 0.59 cm. Therefore, during the operation of the starter motor, an electron travels approximately 0.59 cm along the wire.
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ignore the friction force, and determine the acceleration of the barge when each donkey exerts a forme of 408 N on a cable. m/s
2
The acceleration of the barge is directly proportional to the net force and inversely proportional to the mass. The acceleration of the barge is
816 N / m, where m is the mass of the barge.
The net force on the barge is equal to the force exerted by each donkey, so the net force is 2 * 408 N = 816 N.
The mass of the barge is not given, so we can't calculate the acceleration directly. However, we can say that the acceleration is directly proportional to the net force and inversely proportional to the mass.
If we let the acceleration be represented by the variable a, we can write the following equation:
a = 816 N / m
where m is the mass of the barge.
We can't solve this equation for m, but we can say that the acceleration of the barge is 816 N / m.
In other words, the acceleration of the barge depends on the mass of the barge. If the mass of the barge is larger, the acceleration will be smaller. If the mass of the barge is smaller, the acceleration will be larger.
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Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8×10
11
solar masses, A star orbiting near the galaxy's periphery is 5.9×10
4
light years from its center. (For your calculations, assume that the galaxy's mass is concentrated near its center.) (a) What should the orbital period of that star be? ve yr (b) If its period is 6.5×10
7
years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies. solar masses:
a) The orbital period of that star is 1.10×10 8 yr.
b) The mass of the galaxy is 2.10×10 12 solar masses.
a) Mass of the Milky Way galaxy = 8×10 11 solar masses.
Distance of star from the center of the galaxy, r = 5.9×10 4 light-years.
Force of attraction between the star and the galaxy,
F = GMm/r ²
Here,
M = mass of the galaxy,
m = mass of the star,
r = distance between the star and the galaxy,
G = gravitational constant
Orbital speed,
v = 2πr/T,
where
T is the orbital period of the star
Using the Third Law of Kepler,
T²/R³= 4π²/GM --------(1)
Where
R is the distance of star from the center of the galaxy
T² = (4π²/GM)×R³ = (4π²/GM)(5.9×10 4 × 9.46×10 15 )³ yr ²...[putting R = 5.9×10 4 light-years = 5.9×10 4 × 9.46×10 15 m]T² = (4π²/GM)(2.09×10 41 ) yr ²
T² = 1.23×10 21 (M/M☉) yr ²
On comparing this with the standard formula,
T² = (4π²/GM)R³
We get,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
We know that for the Sun,
M = M☉ and T = 1 year
So,1 year = R³ × 1.51×10 - 8 yr ²
1 year = R³ × 1.51×10 - 8 yr ²
T = (5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 yr
T = 1.10×10 8 yr
(b) We have,
T² = R³ × (M/M☉) × 1.51×10 - 8 yr ²
T² = (6.5×10 7 )² yr ²
R³ = (5.9×10 4 × 9.46×10 15 )³ m ³
On substituting these values, we get
(6.5×10 7 )² yr ²= (5.9×10 4 × 9.46×10 15 )³ × (M/M☉) × 1.51×10 - 8 yr ²
M = (6.5×10 7 )²/[(5.9×10 4 × 9.46×10 15 )³ × 1.51×10 - 8 ] × M☉
M = 2.10×10 12 solar masses.
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A charged particle beam (shot horizontally) moves into a region where there is a constant magnetic field of magnitude 0.00343 T that points straight down. The charged particles in the beam move in a circular path of radius 3.15 cm. If the charged particles in the beam were accelerated through a potential difference of 144 V, determine the charge to mass ratio of the charged particles in the beam. Answer in units of C/kg. 00610.0 points A simple mass spectrometer consists of an accelerating electric potential (so that ions of different mass have different velocities) and a uniform magnetic field (so that different velocity ions have different radii paths). This mass spectrometer is analyzing a beam of singly ionized unknown atoms using the following settings: - the magnetic field is 0.152 T; - the charge of an atom is 1.60218×10
−19
C; - the radius of the orbit is 0.0863 m; and - the potential difference is 179 V. Calculate the mass of the unknown atom. Answer in units of kg
The mass of the unknown atom is 4.65 x 10⁻²³ kg.
The equation for the motion of a charged particle in a magnetic field is given by the equation,
F = Bqv
where:
F = force experienced by the charged particle,
B = magnetic field strength,
q = charge on the particle,
v = velocity of the particle perpendicular to the magnetic field,
r = radius of curvature of the path of the particle, and
m = mass of the charged particle
The beam of charged particles moves in a circular path of radius r = 3.15 cm.
Thus the equation for the radius of the path can be given by,
mv²/r = Bqv
The potential difference of 144 V accelerates the charged particles, which gives them an initial kinetic energy of 144 eV. This can be written as the product of the charge and potential difference,
KE = eΔV
where:
e = charge of an electron and ΔV = potential difference
Thus,
KE = (1.602 x 10⁻¹⁹ C)(144 V)KE = 2.3 x 10⁻¹⁶ Joules
Using the equation of conservation of energy,1/2mv² = eΔVand substituting the value of the velocity of the charged particle from this equation into the first equation, the charge to mass ratio of the charged particles in the beam can be found.
mv²/r = Bqv
Where,
m/e = v/(Br)Charge to mass ratio of the charged particles in the beam can be given by,
e/m = Br²/v
Substitute the given values in the above equation,0.00343 T × (0.0315 m)² = (1.602 x 10⁻¹⁹ C)/(e/m)
Thus,
e/m = 1.7589 x 10¹¹ C/kg
Now, for the second question, the mass of the unknown atom can be found using the equation,
m/e = B²r²/2V
Where,
m = mass of the unknown atom,
e = charge of the unknown atom,
B = magnetic field strength,
r = radius of the path of the unknown atom
V = potential difference
The charge of an atom is 1.60218 x 10⁻¹⁹ C, and the magnetic field is 0.152 T.
The radius of the orbit is 0.0863 m, and the potential difference is 179 V. Substituting these values in the above equation,
m/e = (0.152² x 0.0863²)/(2 x 179)
Thus,
m/e = 2.902 x 10⁻⁴ kg/C
The mass of the unknown atom can be calculated using,
m = e(m/e)
Substituting the known values,
m = (1.602 x 10⁻¹⁹ C)(2.902 x 10⁻⁴ kg/C)
Thus,
m = 4.65 x 10⁻²³ kg
Therefore, the mass of the unknown atom is 4.65 x 10⁻²³ kg.
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Two lenses L1 and L2 of focal lengths 10 cm and 20 cm, are separated by a distance of 80 cm. A 5 cm tall is placed 14 cm from the leftmost lens (L1). Find the location of the final image. Find the focusing power of the system (in diopters).
Focal length of L1, f1 = -10 cm Focal length of L2, f2 = -20 cm Distance between the lenses, d = 80 cm Height of the object, h1 = 5 cm Distance of the object from the first lens, u1 = 14 cm. The location of the final image is 92 cm from the second lens (L2).The focusing power of the system is -86.2069 diopters.
The first lens is concave and the second lens is convex. The first lens will form a virtual, erect and diminished image at a distance, v1, given by the lens formula,1/f1 = 1/u1 + 1/v1 => 1/v1 = 1/f1 - 1/u1=> 1/v1 = - 1/10 - 1/14 => 1/v1 = -24/140=> v1 = -5.83 cm (negative sign shows that it is a virtual image)
The second lens will form a real, inverted and diminished image of the virtual image at a distance, v2, given by the lens formula,1/f2 = 1/u2 + 1/v2 => 1/v2 = 1/f2 - 1/u2=> 1/v2 = - 1/20 - 1/-5.83 => 1/v2 = 0.0833=> v2 = 12 cm (positive sign shows that it is a real image)The final image is formed at a distance of (d + v2) from the second lens L2,i.e. v = 80 + 12 = 92 cm
The magnification produced by the first lens, m1, is given by,m1 = v1 / u1 = -5.83 / 14 = -0.4179The magnification produced by the second lens, m2, is given by,m2 = v2 / v1 = -12 / -5.83 = 2.06The total magnification, m, produced by the system is the product of m1 and m2,m = m1 * m2 = -0.4179 * 2.06 = -0.861The focusing power, F, of the system is given by, F = 1/f => F = 1/-0.0116 => F = - 86.2069 D
The location of the final image is 92 cm from the second lens (L2).The focusing power of the system is -86.2069 diopters.
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