Answer: 5cm
Explanation: Since the radius of curvature is 10cm, the focal length of the mirror (f) is
f = r/2
Where r is the radius of curvature.
Since r = 10cm, f = 10/2 = 5cm.
To produce a parallel light rays of a flash light, it means the the image will be at infinity this making the image distance to be infinite.
From the mirror formulae
1/u + 1/v = 1/f
Where u = object distance =?, v = image distance = infinity and f = focal length = 5cm.
Let us substitute the parameters, we have that
1/u + 1/∞ = 1/5
1/∞ = 0
Hence 1/u = 1/5
u = 5cm.
Hence the object needs to be placed nearly 5cm to the mirror
Final answer:
The light bulb should be placed at the focal length of the concave mirror, which is 5 cm, to produce near-parallel light rays in a flashlight.
Explanation:
The distance between the light bulb and a concave mirror to produce a beam of near-parallel light rays in a flashlight is most nearly the focal length of the mirror. Since the mirror's radius of curvature (R) is 10 cm, using the formula R = 2f, we can calculate the focal length (f). Dividing the radius of curvature by 2, we get f = R/2 = 10 cm / 2 = 5 cm. Therefore, the light bulb should be placed approximately 5 cm from the mirror to achieve near-parallel rays.
g Two hollow conducting spheres (radius ????) with a uniformly distributed charge are placed a distance ???? apart center to center. A thin wire with a switch ???? is connected to the surface of each sphere. The switch is initially open. a. What is the potential between points ???? and ????? b. If the switch is then closed, what is the charge on each sphere at time ???? → [infinity]? c. What is the potential between points ???? and ???? after the sphere reaches its steady state?
The given question is incomplete. The complete question is as follows.
Two hollow conducting spheres (radius r) with a uniformly distributed charge are placed a distance d apart center to center. A thin wire with a switch S is connected to the surface of each sphere. The switch is initially open.
a. What is the potential between points a and b?
b. If the switch is then closed, what is the charge on each sphere at time [tex]t \rightarrow \infty[/tex].
c. What is the potential between points a and b after the sphere reaches its steady state?
Explanation:
(a) In order to bring a positive test charge from infinity to a point 'a', the work done is equal to the potential energy of the charge at point 'a'.
Hence, [tex]V_{a} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{a}[/tex]
Now, work done in bringing a positive test charge from infinity to point 'b' is equal to the potential energy of the charge at point 'b'.
[tex]V_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{-q}{b}[/tex]
[tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]
Therefore, the potential between points a and b is as follows.
[tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]
(b) As the spheres are connected through a conducting wire then charges will flow from one sphere to another unless and until the charge on both the sphere will become equal. In this case, it is equal to zero.
(c) Since, the charge of both the spheres is equal to zero so, no work is necessary to bring another charge to a and b. Therefore, potential difference between the points will also become equal to zero.
A proton is circling the Earth above the magnetic equator, where Earth’s magnetic field is directed horizontally north and has a magnitude of 4.00 × 10–8 T. If the proton is moving at a speed of 2.7 × 107 m/s, how far above the surface of the Earth is the proton
Answer:
[tex]6.65\times 10^5 m[/tex]
Explanation:
We are given that
Magnetic field=B=[tex]4\times 10^{-8} T[/tex]
[tex]v=2.7\times 10^7 m/s[/tex]
We have to find the height of proton from the surface of the Earth.
Mass of proton,[tex]m_p=1.67\times 10^{-27} kg[/tex]
Charge on proton,[tex]q=1.6\times 10^{-19} C[/tex]
Radius of Earth, r=[tex]6.38\times 10^6 m[/tex]
Centripetal force due to rotation of proton=[tex]\frac{mv^2}{r+h}[/tex]
Magnetic force,F=[tex]qvB[/tex]
[tex]\frac{mv^2}{r+h}=qvB[/tex]
[tex]\frac{mv}{r+h}=qB[/tex]
Substitute the values
[tex]\frac{1.67\times 10^{-27}\times (2.7\times 10^7)}{6.38\times 10^6+h}=1.6\times 10^{-19}\times 4\times 10^{-8}[/tex]
[tex]6.38\times10^6+h=\frac{1.67\times 10^{-27}\times 2.7\times 10^7}{1.6\times 10^{-19}\times 4\times 10^{-8}}[/tex]
[tex]6.38\times10^6+h=7.045\times 10^6[/tex]
[tex]h=7.045\times 10^6-6.38\times 10^6[/tex]
[tex]h=0.665\times 10^6=6.65\times 10^5 m[/tex]
The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be the fundamental frequency on a piece of the same string that is twice as long and has four times the tension
The fundamental frequency of the string will remain at 250 Hz, even when the string is doubled in length and the tension is quadrupled, due to the relationship between string length, tension, and frequency.
Explanation:When a transverse standing wave on a string is vibrating at its fundamental frequency, the length of the string and the tension within it play crucial roles in determining that frequency. The fundamental frequency (f) is given by f = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the linear mass density of the string.
In the original scenario, the string vibrates at a fundamental frequency of 250 Hz. If the same piece of string is made twice as long and the tension is increased fourfold, the new fundamental frequency F' can be found using the formula mentioned above.
Considering the string is now twice as long (L' = 2L), and the tension is four times greater (T' = 4T), we can substitute these into the original equation for fundamental frequency: f' = (1/2L') * sqrt(T'/μ). By inserting these new values, we can determine that the new fundamental frequency f' will be the same as the original frequency f, because the doubling of the string length and the quadrupling of the tension will have cancelling effects.
Final answer:
The fundamental frequency of a string that is twice as long and with four times the tension would be half the original frequency; therefore, the new fundamental frequency would be 125 Hz.
Explanation:
The student is asking about how the fundamental frequency of a string changes when its length is doubled and the tension is increased by four times. The fundamental frequency f of a string fixed at both ends can be described by the equation f = √(T/μ)/(2L), where T is the tension, μ (mu) is the mass per unit length, and L is the length of the string. When we double the length of the string, the new length is 2L. When we quadruple the tension, the new tension is 4T. Substituting these values into the equation, the new frequency f' becomes f' = √(4T/μ)/(2 × 2L) = √(T/μ)/2L = f/2.
Therefore, the new fundamental frequency of the longer string with four times the tension is half the original frequency, which is 125 Hz in this case.
The force that generates the heat and light produced by the sun and other stars is 1. the electromagnetic force. 2. the weak force. 3. the strong force. 4. the gravitational force.
The electromagnetic force generates the heat and light produced by the sun and other stars.
Explanation:The force that generates the heat and light produced by the sun and other stars is the electromagnetic force. This force is responsible for holding atoms together and producing electromagnetic radiation, which includes heat and light. It is much stronger compared to the weak force and gravity.
You have a battery marked " 6.00 V ." When you draw a current of 0.361 A from it, the potential difference between its terminals is 5.07 V . What is the potential difference when you draw 0.591 A
Answer:
Explanation:
Battery voltage is 6V
A current of 0.361A is draw the voltage reduces to, 5.07V
This shows that the appliances resistance that draws the currents is
Using KVL
The battery has an internal resistance r
V=Vr+Va
Vr is internal resistance voltage
Va is appliance voltage
6=5.07+Va
Va=6-5.07
Va=0.93
Using ohms law to the resistance of the appliance
Va=iR
R=Va/i
R=0.93/0.361
R=2.58ohms
Then if the circuit draws a current of 0.591A
Then the voltage across the load is
V=iR
Va=0.591×2.58
Va=1.52V
Then the voltage drop at the internal resistance is
V=Vr+Va
Vr=V-Va
Vr=6-1.52
Vr=4.48V
Answer:
V = 4.48 V
Explanation:
• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:
[tex]V_{rint} = I* r_{int}[/tex]
• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:
[tex]V = V_{b} - V_{rint} = 5.07 V = 6.00 V - 0.361 A * r_{int}[/tex]
• We can solve for rint, as follows:
[tex]r_{int} =\frac{V_{b}- V_{rint} }{I} = \frac{6.00 V - 5.07 V}{0.361A} = 2.58 \Omega[/tex]
• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:
[tex]V = V_{b} - V_{rint} = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V[/tex]
• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.
"Comparing microwaves and visible light, which of the following is true? 1. Microwaves have higher frequency, same speed, and longer wavelength than visible light. 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light. 3. Microwaves have lower frequency, slower speed, and longer wavelength than visible light. 4. Microwaves have lower frequency, faster speed, and shorter wavelength than visible light."
Answer:
2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.
Explanation:
Microwaves are a form of electromagnetic radiation. Most people are familiar with this type of waves because they are used in microwave ovens. When compared to visible light, microwaves have lower frequency, same speed and longer wavelength than visible light. The prefix "micro" is used to indicate that microwaves are smaller (shorter wavelengths) than radio waves.
"The correct option is 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.
To understand why this option is correct, let's consider the relationship between frequency, wavelength, and speed for electromagnetic waves, which is given by the equation:
[tex]\[ c = f \times \lambda \][/tex]
1. Speed of light (in a vacuum) is constant for all electromagnetic waves, including microwaves and visible light. Therefore, the speed of microwaves and visible light is the same.
2. Microwaves have a lower frequency than visible light. The frequency of microwaves typically ranges from 300 MHz to 300 GHz, while the frequency of visible light ranges from approximately 430 THz to 750 THz. Since microwaves have a lower frequency, they also have a longer wavelength according to the equation [tex]\( c = f \times \lambda \)[/tex].
3. Since the speed of all electromagnetic waves is the same in a vacuum, and microwaves have a lower frequency, they must have a longer wavelength to maintain the constant speed. This is consistent with the relationship , where a lower frequency requires a longer wavelength to keep the speed constant.
In summary, microwaves have a lower frequency and longer wavelength than visible light, but they travel at the same speed. This makes option 2 the correct choice."
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.7 m/s at the top of the arc
Answer:
1) [tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]
Explanation:
1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:
[tex]K_{A} = K_{B} + U_{g,B}[/tex]
[tex]\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}[/tex]
[tex]\frac{1}{2} \cdot v_{A}^{2} = \frac{1}{2} \cdot v_{B}^{2} + g \cdot h_{B}[/tex]
The minimum speed is obtained herein:
[tex]v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}[/tex]
[tex]v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}[/tex]
[tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]
The minimum speed he need at launch point is [tex]8.27m/s[/tex]
Energy conservation :Energy is neither be created nor be destroyed just change into one form to another form.
[tex]\frac{1}{2}mv^{2} =\frac{1}{2}mv_{a}^{2} +mgh\\ \\v=\sqrt{v_{a}^{2}+2gh }[/tex]
Where,
[tex]v[/tex] is velocity at launch.[tex]v_{a}[/tex] is velocity at the top of the arc[tex]g[/tex] is gravitational acceleration, [tex]g=9.8m/s^{2}[/tex][tex]h[/tex] is height of center of mass.Given that, [tex]v_{a}=6.7m/s,h=1.2m,g=9.8m/s^{2}[/tex]
Substitute all values in above relation.
[tex]v=\sqrt{(6.7)^{2}+2*9.8*1.2 } \\\\v=\sqrt{68.41}=8.27m/s[/tex]
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Two radio antennas A and B radiate in phase. Antenna B is a distance of 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. What is the longest wavelength for which there will be destructive interference at point Q?
Answer:
240 m
120 m
Explanation:
d = Path difference = 120 m
For destructive interference
Path difference
[tex]d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m[/tex]
The longest wavelength is 240 m
For constructive interference
[tex]d=\lambda\\\Rightarrow 120\ m=\lambda[/tex]
The longest wavelength is 120 m
Final answer:
The longest wavelength for destructive interference at point Q, where the path difference is 120 m, is 240 m.
Explanation:
The longest wavelength for which there will be destructive interference at point Q can be found by considering the path difference between the waves from the two antennas at point Q. The waves must differ by an odd multiple of ½ wavelengths for destructive interference to occur. The distance from antenna A to point Q is 120 m + 40 m = 160 m, and the distance from antenna B to point Q is 40 m. Therefore, the path difference is 120 m.
For the first occurrence of destructive interference, the path difference should be ½ wavelength, so we calculate the longest wavelength (λ) as:
½ λ = 120 m
λ = 240 m
This is the longest wavelength that causes destructive interference at point Q.
Problem 7: In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.
Answer:
Complete question
In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.
a. What electric field strength, in volts per mater, is needed to select a speed of 4.2 x 10^6 m/s?
b. What is the voltage, in kilovolts, between the plates if they are separated by 0.95 cm?
Explanation:
Given that,
magnetic field B = 0.095T
Speed of particle v = 4.2 ×10^6m/s
Separation between plate d = 0.95cm
d = 0.95/100 = 0.0095m
a. Using the mass spectrometer velocity selector relationship between the electric field and magnetic field.
v = E/B
Where
v is the speed selector
B is magnetic field
E is electric field
Therefore, E = vB
E = 4.2 × 10^6 × 0.095
E = 0.399× 10^6
E = 3.99 × 10^5 V/m
b. Voltage?
The relationship between electric field and potential difference between the two plates is given as
V = Ed
V = 3.99 × 10^5 × 0.0095
V = 3790.5 V
To kV, 1kV = 1000V
Then, V = 3.7905kV
V ≈ 3.791 kV
The electric field strength needed in a mass spectrometer to select a velocity of 4.00 × 106 m/s with a 0.100-T magnetic field is 400,000 N/C. The voltage required across plates separated by 1.00 cm is 4,000 V.
Explanation:In a velocity selector within a mass spectrometer, a charged particle remains undeflected when the electric force equals the magnetic force. This condition is given by qE = qvB, where q is the charge, E is the electric field strength, v is the velocity of the particle, and B is the magnetic field strength. The question asks for the electric field strength needed to select particles with a velocity of 4.00 × 106 m/s when subjected to a 0.100-T magnetic field. Using the condition for balance, E = vB, we can substitute the given values to find E = 4.00 × 106 m/s × 0.100 T = 400,000 N/C. For part (b), voltage (V) can be determined by the relation V = Ed, where d is the separation between the plates. For a plate separation of 1.00 cm, or 0.01 m, the voltage is V = 400,000 N/C × 0.01 m = 4,000 V.
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65 T.
What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6 1 × 107 m/s?
Answer:
10581.59 V
Explanation:
We are given that
Magnetic field=B=0.65 T
Speed of electron=[tex]v=6.1\times 10^7m/s[/tex]
Charge on electron, [tex]q=e=1.6\times 10^{-19} C[/tex]
Mass of electron,[tex]m_e=9.1\times 10^{-31} kg[/tex]
We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.
[tex]V=\frac{v^2m_e}{2e}[/tex]
Where V=Potential difference
[tex]m_e=[/tex]Mass of electron
v=Velocity of electron
Using the formula
[tex]V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}[/tex]
[tex]V=10581.59 V[/tex]
Hence, the potential difference=10581.59 V
Final answer:
To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, the potential difference required in the first part of the experiment is approximately 88.6 volts.
Explanation:
To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, we need to calculate the potential difference required in the first part of the experiment. The formula for the potential difference is given by:
V = (1/2)m*(v^2)/(q * B)
Where V is the potential difference, m is the mass of the electron (9.11 × 10^-31 kg), v is the velocity of the electron (6.1 × 10^7 m/s), q is the charge of the electron (-1.6 × 10^-19 C), and B is the magnetic field strength (0.65 T).
Plugging in the values into the formula, we get:
V = (1/2)(9.11 × 10^-31 kg)(6.1 × 10^7 m/s)^2/(-1.6 × 10^-19 C)(0.65 T)
Simplifying the expression, we find that the potential difference required is approximately 88.6 volts.
Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits two wires simultaneously with its wings
Answer:
The reason the bird is not electrocuted is due to some facts about circuit:
1. Completeness of circuit- This circuit needs to be
complete in order for current to flow. The bird standing on only one wire has not completed the circuit.
2. A potential difference: Another factor deciding
the direction of flow of current is (electric)
potential. Current always flows from a higher
potential to a lower potential. In other words it
can be said that electrons flow from lower
potential to higher one. (the direction of electric
current is opposite to that of the electrons). So
we need the potential difference for current to
flow. The bird standing on only one wire has no potential difference.
3. Path of least Resistance- Factor that decides
the path a current will flow in case of parallel
paths is the (electric) resistance offered by the
path. Current will always flow in the path that
offers least resistance. The leg of a bird has high resistance.
Explanation:
It has no potential difference as both the legs of bird are touching the same wire at same constant potential. ... If the bird would touch the ground while sitting on the wire or flap its wings and touch another electric wire with a different voltage, then it would get shocked and likely die by electrocution.
while in the other hand, the Bird that touches two wires with it wings at the same time will get electrocuted because it has completed a circuit and the its feathers created a potential difference .
A bird is not electrocuted on a high-voltage power line because it's not completing a circuit. By contrast, if a large bird touches two wires simultaneously, it creates a closed circuit, allowing electricity to flow through its body, leading to electrocution. Circuit breakers serve to prevent excess current flow.
Explanation:A bird sitting on a high-voltage power line does not get electrocuted because it is not completing a circuit. The bird is not touching the ground or another line, so the electricity stays within the conducting wire and doesn't pass through the bird. The bird is safe as long as it touches only one wire.
However, in contrasting scenarios, a larger bird could span the distance between two wires with its wings. If this happens, the bird would close the circuit, allowing the electricity to pass through its body, which would lead to electrocution. This is due to the significance of voltage difference between two given points to create an electric shock, as there is a potential difference between the two wires.
Also, the role of circuit breakers is important in the context of high voltage power supply. Circuit breakers prevent excess current flow and accidental contact with the line. They interrupt the electricity flow anytime there's a fault, safeguarding anyone who may come in contact with the cable.
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Calculate the noise voltage spectrum Su(f) on a paraller RC circuit with R=1 kOhm and C=1 nF. (Don't forget to provide the unit!) (3 points) - How much it is at 1 Hz ?
Answer: 10^-3 V^2/Hz
Explanation:
1 Hz:
Su(f) = No * |H(f)|^2
= 10^-3 * 1/(1+(2*pi*f*R*C)^2)
= 10^-3 V^2/Hz
Ocean waves are traveling to the east at 4.9 m/s with a distance of 23 m between crests. (b) With what frequency do the waves hit the front of a boat when the boat is moving westward at 1.1 m/s
Answer:
0.26086 Hz
Explanation:
solution:
The relative speed of the wave is:
v = 4.9+1.1 = 6 m/s
the frequency is:
f = v/λ
= 6/23
= 0.26086 Hz
A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 1.6 m/s. The block slides and stops at a distance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were increased by a factor of 3.5
Answer:
19.6 m
Explanation:
The work-energy theorem applies here,
The theorem states that the change in momentum of a particle between two points is equal to the work done in moving the force between the two distance.
ΔK.E = W
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Kinetic energy = (1/2)(m)(v²)
m = 3.5 kg, v = 1.6 m/s
Final kinetic energy = 0 J, since the block of mass comes to rest.
Initial kinetic energy = (1/2)(3.5)(1.6²) = 4.48 J
ΔK.E = - 4.48 J
The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement.
W = - Fd = - 1.6 F
ΔK.E = W
- 4.48 = - 1.6 F
F = 2.8 N
when the initial velocity is increased by a factor 3.5,
v = 1.6×3.5 = 5.6 m/s
Final kinetic energy = 0 J, since the block of mass comes to rest.
Initial kinetic energy = (1/2)(3.5)(5.6²) = 4.48 J
ΔK.E = - 54.88 J
The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement. The frictional force is the same as above.
W = - Fd = - (2.8) d
ΔK.E = W
- 54.88 = - 2.8 d
d = 19.6 m
Compared to 1.6 m, 19.6 m is an increase by a factor 12.25.
If you pay him $200 up front, Freddie promises to paint your garage. But you'd be crazy to pay him anything up front because Freddie lies all the time, and every dime he gets his hands on he uses to feed his methamphetamine habit.a. False dichotomyb. Accident.c. Argument against the person, abusived. False cause.e. Argument against the person, circumstantial.
Answer: e
Explanation:
Argument against the person, circumstantial.
A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P). Choose the planet to be your system. The work done by the force of gravity during this movement is:
Answer:
Zero work done,since the body isn't acting against or by gravity.
Explanation:
Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.
Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.
The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.
If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RC constant is acceptable, and given that the flash is driven by a 600-μF capacitor, what is the resistance in the flash tube?
Explanation:
The given data is as follows.
Speed of the bullet (v) = 500 m/s
Distance during one RC time constant (d) = 1 mm = [tex]1 \times 10^{-3} m[/tex]
Capacitance (C) = 600 [tex]\mu F[/tex] = [tex]600 \times 10^{-6} F[/tex]
Hence, formula for speed of the bullet is as follows.
v = [tex]\frac{d}{t}[/tex]
or, t = [tex]\frac{d}{v}[/tex]
Time constant for RC circuit is as follows.
t = RC
R = [tex]\frac{t}{C}[/tex]
= [tex]\frac{d}{vC}[/tex]
= [tex]\frac{1 \times 10^{-3}}{500 \times 600 \times 10^{-6}}[/tex]
= [tex]3.33 \times 10^{-3} ohm[/tex]
Thus, we can conclude that resistance in the flash tube is [tex]3.33 \times 10^{-3} ohm[/tex].
A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)
Approximately 1.6 million joules of work is required to pump the liquid out of the spherical tank through a 1 m long spout at the top.
Explanation:The work done to pump the liquid out of a tank can be calculated using the formula W = ρgVh, where ρ is the density of the liquid, g is the acceleration due to gravity, V is the volume of the liquid, and h is the height up to which the liquid is pumped. Since the tank is spherical and half full, the volume of the liquid is ½(4/3πr³), or 2πr³. Substituting the given values: ρ = 900 kg/m³, g = 9.8 m/s², r = 3 m, and h is the radius of the sphere plus the length of the spout (3 m + 1 m = 4 m), we get W = 900 kg/m³ * 9.8 m/s² * 2π(3 m)³ * 4 m ≈ 1.6 * 10⁶ J. Therefore, approximately 1.6 million joules of work is required to pump the liquid out of the tank.
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During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial velocity v0 = 26 m/s at an angle θ = 17° above horizontal. Let the origin of the Cartesian coordinate system be the ballʼs position at impact. Air resistance may be ignored throughout this problem.
a) express the magnitude of the ball's initial horizontal velocity, v0x, in terms of v0 and theta.
b) express the magnitude of the ball's inital vertical velocity, v0y, in terms of v0 and theta.
c) find the ball's maximum vertical height, hmax, in meters above ground.
d) create an expression in terms of v0, theta, and g for the time (tmax) it takes the ball to travel to its maximum vertical height.
e) calculate the horizontal distance, xmax, in meters the ball has traveled when it returns to ground level.
Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,
c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m
Explanation:
A)
The velocity v0 is at angle θ to the horizontal.
The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.
With vo as the hypotenus, vox as the adjacent and voy as the opposite.
To get vox, we relate vo and vox ( hypotenus and adjacent)
From trigonometry
Cos θ relates hypotenus and adjacent, hence we have that
Cos θ = vox/vo
vox = vo × cos θ
B)
To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).
According to trigonometry, sin θ relates hypotenus and opposite, hence we have that
Sin θ = voy/vo
voy =vo× sin θ
C)
The formulae for the maximum height of a projectile motion is given as
H = vo² (sin θ)²/2g
Where g = acceleration due to gravity = 9.8 m/s²
By substituting the parameters, we have that
H = 26² × (sin 17)²/2(9.8)
H = 676 × 0.0854/19.6
H = 57.7304/ 19.6
H = 2.94 m
D)
This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g
From the equation of motion
vy = voy - gt
0 = voy - gt
But voy = vo sinθ
0 = vo sinθ - gt
gt = vo sinθ
t = vo sinθ / g
E)
The horizontal distance covered formulae is given by
R = u² sin2θ/g
R = 26² × sin 2(17)/9.8
R = 676 × sin 34/ 9.8
R = 378.014/ 9.8
R = 38.57 m
The correct Answer is:
a) vox = vo × cos θ, b) voy =vo× sin θ,c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 mA) When The velocity v0 is at angle θ to the horizontal.When The horizontal component of vo (vox), vo, and also that the vertical component of vo (voy) all form a right angle triangle.Although when With vo as the hypotenuse, vox as the adjacent and voy as the opposite.Then To get vox, we relate vo and also that vox ( hypotenuse and adjacent)When From trigonometryAlso, Cos θ relates hypotenuse and adjacent, hence we have thatThen Cos θ = vox/voThen vox = vo × cos θB) When To get the vertical component of vo, we relate to and also voy ( hypotenuse and also opposite).According to trigonometry, sin θ relates to the hypotenuse and also that opposite, hence we have thatThen Sin θ = voy/voThen voy =vo× sin θ
C) When The formulae for the maximum height of a projectile motion is given asH = vo² (sin θ)²/2gWhere that g = acceleration due to gravity = 9.8 m/s²By substituting the parameters, we have thatThen H = 26² × (sin 17)²/2(9.8)Then H = 676 × 0.0854/19.6Then H = 57.7304/ 19.6Then H = 2.94 m
D) When This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g From the equation of motionvy = voy - gt0 = voy - gtBut voy = vo sinθ0 = vo sinθ - gtgt = vo sinθt = vo sinθ / g
E) When The horizontal distance covered formulae is given byR = u² sin2θ/gR = 26² × sin 2(17)/9.8R = 676 × sin 34/ 9.8R = 378.014/ 9.8R = 38.57 m
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An isolated charged insulating solid sphere with radius of 20 cm is carrying charge of -4.8 x 10-16 A solid sphere is surrounded by a hollow charged to +4.8 x 10-16 C conducting sphere with inner radius of r; = 40 cm and outer radius ro = 50 cm Find an Electric field at
a) r= 30 cm
b) = 45 cm
c) = 60 cm Use following constants: K = 8.99 x 109 Nm2/C2, Ep = 8.854 x 10-12 C2/Nm2
Answer:
a) E = -4.8 10⁻⁵ N / C , b) E = 0 , c) E = 0
Explanation:
For this exercise let's use Gauss's law
Ф = E. dA = [tex]q_{int}[/tex] / ε₀
As a Gaussian surface we use a sphere, whereby the electric field lines are parallel to the normal area, and the scalar product is reduced to the algebraic product
a) the field for r = 30 cm
At this point we place our Gaussian surface and see what charge there is inside, which is the charge of the solid sphere (r> 20cm), the charge on the outside does not contribute to the flow
E = q_{int} / A ε₀
q_{int} = -4.8 10-16 C
The area of a sphere is
A = 4π R²
We replace
E = -4.8 10⁻¹⁶ / 4π 0.30² 8.85 10⁻¹²
E = -4.8 10⁻⁵ N / C
b) r = 45 cm
This point is inside the spherical conductor shell, as in an electric conductor in electrostatic equilibrium the charges are outside inside the shell there is no charge for which the field is zero
E = 0
c) R = 60 cm
This part is outside the two surfaces
The chare inside is
q_{int} = -4.8 10⁻¹⁶ + 4.8 10⁻¹⁶
q_{int} = 0
Therefore the electric field is
E = 0
Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum dark fringe appears 3 mm from the central maximum. What is the width of the slit?
Answer:
Width of the slit will be equal to 1.47 mm
Explanation:
We have given wavelength of the light [tex]\lambda =550nm=550\times 10^{-9}m[/tex]
Distance D = 8 m
Distance between first minimum dark fringe and the central maximum is 2 mm
So [tex]x=3\times 10^{-3}m[/tex]
We have to find the width of the slit
For the first order wavelength is equal to [tex]\lambda =\frac{x}{D}\times a[/tex], here a width of slit
So [tex]a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm[/tex]
So width of the slit will be equal to 1.47 mm
In a high-performance computing system 100 CPU chips, each dissipating 25 W, are attached to one surface of a 200 mm by 200 mm copper heat spreader. The copper plate is cooled on its opposite surface by water, which is tripped at the leading edge of the copper so that the boundary layer is turbulent throughout. The plate may be assumed to be isothermal due to the high thermal conductivity of the copper. The water velocity and temperature are 2 m/s and 17˚C. What is the temperature of the copper plate?
Given Information:
Velocity of water flow = 2 m/s
Temperature of water = 17° C
Heat dissipation = 2500 W
Area of copper plate = 0.04 m²
Required Information:
Temperature of copper plate = ?
Answer:
[tex]T_{p} = 27[/tex]° [tex]C[/tex]
Explanation:
Each chip dissipates 25 W so 100 chips will dissipate 25*100 = 2500 W
Area of copper plate = 0.2*0.2 = 0.04 m²
According to the convection rate equation
[tex]T_{p}= T_{w} + \frac{q}{hA}[/tex]
Where Tp is the temperature of copper plate, Tw is the temperature of water, q is the the heat dissipation of chips, A is the area of copper plate and h is the convection coefficient
The convection coefficient is given by turbulent flow correlation
[tex]h = Nu_{L}(k/L) =0.037Re_{L}^{4/5}P_{r}^{1/3}(k/L)[/tex]
Where Nu is Nusselt number, Re is Reynolds number, Pr = 5.2 is Prandtl number and k = 0.620 W/m.K
[tex]Re_{L}= uL/v[/tex]
Where u = 2 m/s and L = 0.2 m and v = 0.96x10⁻⁶m² /s
[tex]Re_{L}= 2*0.2/0.96x10^{-6}[/tex]
[tex]Re_{L}= 416666.66[/tex]
[tex]h = 0.037(416666.66)^{4/5}(5.2)^{1/3}(0.620/0.2)[/tex]
[tex]h = 6223.89[/tex] [tex]W/m^{2}K[/tex]
[tex]T_{p}= 17 + \frac{2500}{(6223.89)0.04}[/tex]
[tex]T_{p} = 27[/tex]° [tex]C[/tex]
Therefore, the temperature of the copper plate is 27° C
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 95.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 21.0 kg.
(c) Calculate the acceleration.
m/s2
(d) What would the acceleration be if friction is 20.0 N?
Final answer:
The acceleration of the child in the wagon can be calculated using Newton's Second Law of Motion. The sum of the forces acting on the system is calculated by adding the forces exerted by the children and subtracting the force of friction. Using the given values, the acceleration can be determined by dividing the sum of the forces by the mass of the system. If the force of friction is increased to 20.0 N, the acceleration can be recalculated using the new value.
Explanation:
To calculate the acceleration of the child in the wagon, we can use Newton's Second Law of Motion:
ΣF = ma
Where ΣF is the sum of all the forces acting on the system, m is the mass of the system, and a is the acceleration.
In this case, the forces acting on the system are the forces exerted by the two children and the force of friction.
Using the given values:
Force1 = 75.0 N
Force2 = 95.0 N
Friction = 12.0 N
Mass = 21.0 kg
The sum of the forces acting on the system is:
ΣF = Force1 + Force2 - Friction
Substituting the values:
ΣF = 75.0 N + 95.0 N - 12.0 N
ΣF = 158.0 N
Now we can calculate the acceleration:
a = ΣF / m
Substituting the values:
a = 158.0 N / 21.0 kg
a = 7.52 m/s²
Therefore, the acceleration of the child in the wagon is 7.52 m/s².
(d) If friction is 20.0 N, we can recalculate the acceleration using the new value:
ΣF = 75.0 N + 95.0 N - 20.0 N
ΣF = 150.0 N
a = ΣF / m
a = 150.0 N / 21.0 kg
a = 7.14 m/s²
Therefore, if friction is 20.0 N, the acceleration of the child in the wagon would be 7.14 m/s².
A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiving end absorbs 10 MVA at 33 kV. Assuming a short line, calculate: (a) the ABCD parameters, (b) the sending-end voltage for a load power factor of 0.9 lagging, and (c) the sending-end voltage for a load power factor of 0.9 leading.
Answer:
(a) With a short line, the A,B,C,D parameters are:
A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu
(b) The sending-end voltage for 0.9 lagging power factor is 35.96 [tex]KV_{LL}[/tex]
(c) The sending-end voltage for 0.9 leading power factor is 33.40 [tex]KV_{LL}[/tex]
Explanation:
(a)
Considering the short transition line diagram.
Apply kirchoff's voltage law to the short transmission line.
Write the equation showing the relations between the sending end and the receiving end quantities.
Compare the line equations with the A,B,C,D parameter equations.
(b)
Determine the receiving-end current for 0.9 lagging power factor.
Determine the line-to-neutral receiving end voltage.
Determine the sending end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
(c)
Determine the receiving-end current for 0.9 leading power factor.
Determine the sending-end voltage of the short transition line.
Determine the line-to-line sending end voltage which is the sending end voltage.
Two point charges Q1 = +4.10 nC and Q2 = −2.40 nC are separated by 55.0 cm.(a) What is the electric potential at a point midway between the charges? 212.7 Incorrect: Your answer is incorrect. Use the expression for the electric potential from each point charge to find the electric potential at the midpoint between the two charges. V(b) What is the potential energy of the pair of charges? 3.22E-7 Incorrect?
Answer:
a. 55.6v
b. [tex]1.61*10^{-7}J[/tex]
Explanation:
Data given
charge 1=+4.10nC
charge 2 =-2.40nC
distance, r= 55cm =0.55m
a. the electric potential at the mid point is the sum of the potential due to individual charge.
The electric potential is expressed as
[tex]V=\frac{kq}{r}\\[/tex]
since we are interested in the electric potential at the mid point, we have
[tex]V=\frac{kq_1}{r/2}-\frac{kq_2}{r/2}\\ V=\frac{2k}{r}(q_1-q_2)\\ V=\frac{2*9*10^9}{0.55}(4.1-2.4)*10^{-9}\\ V=55.6v[/tex]
Hence the electric potential at the mid-point is 55.6v
b. to calculate the potential energy, we use the formula below
[tex]U=\frac{kq_1q_2}{r} \\U=\frac{9*10^9 *4.10*10^{-9}*2.4*10^{-9}}{0.55}\\ U=1.61*10^{-7}J[/tex]
Two plates with area A are held a distance d apart and have a net charge +Q, and -Q, respectively. Assume that all the charge is uniformly distributed on the inner surfaces of the plates.
The left plate has charge -Q, the right plate has charge +Q, separated by distance d.
1) Find the charge density on the plates.
2) Find the electric potential difference between the plates.
3) Show that the capacitance of the enlarged plates in this case is the same as the capacitance in a case where
Answer:
Explanation:
1 )
Charge density of left plate
= - Q / A
Charge density of right plate
= + Q / A
2 )
capacitance c = ε₀ A / d
potential difference = charge / capacitance
= Q / [ ε₀ A / d ]
= Q d / ε₀ A
Final answer:
The charge density on each plate is ±Q/A, the electric potential difference between the plates is calculated using the electric field and the separation distance, and the capacitance C=ε₀A/d demonstrates that the size of the plates does not affect their capacitance as long as their proportion remains constant.
Explanation:
When considering two parallel plates each with area A and charges of +Q and -Q respectively, separated by a distance d, we can address the posed questions systematically.
Finding the Charge Density on the Plates
The surface charge density σ is defined as charge per unit area. Given the total charge +Q or -Q and the area A of each plate, the charge density on each plate is σ = ±Q/A. This is a direct result of the uniform distribution assumption of the charges across the plates.
Finding the Electric Potential Difference Between the Plates
The electric field E created between the plates by the charge distribution is uniform and can be represented as E = σ/ε₀, where σ is the surface charge density and ε₀ is the vacuum permittivity. Consequently, the electric potential difference V between the plates can be derived from the relation V = Ed, linking the electric field and the separation of the plates.
Demonstrating the Capacitance of Enlarged Plates Remains Constant
The capacitance C of a parallel-plate capacitor is given by C = ε₀A/d, which is independent of the charge on the plates. This formula illustrates that the capacitance is solely dependent on the physical characteristics of the capacitor (i.e., the area of the plates A, the distance between them d, and the permittivity of free space ε₀), and does not change with the amount of charge nor with the size of the plates as long as their proportional relationship remains constant.
When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m . You then use your thumb to partially cover the output opening so that its diameter is reduced to 10 mm.
Part A
How high does the water rise now? Ignore drag and assume that the smaller opening you create with your thumb is circular.
Express your answer with the appropriate units
h=
Answer: h = 0.52m
Explanation:
Using the equation of out flow;
A1 × V1 = A2 ×V2
Where A1 = area of the first nozzle
A2 = area of the second nozzle
V1= velocity of flow out from the first nozzle
V2 = velocity of flow out from 2nd nozzle
But AV= area of nozzle × velocity of water = volume of water per second(m³/s).
Now we can set A×V = Area of nozzle × height of rise.
Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)
D1 = 20mm= 0.02m; h1 = 0.13m
D2 = 10mm = 0.01m; h2= ?
h2 = π(D1/2)²× h1 /π(D2/2)²
h2 = (0.02/2)² × 0.13/(0.01/2)²
= (0.01)² ×0.13 /(0.005)²
= 1.3 × 10^-5/(5 × 10^-3)²
= 1.3 × 10^-5/25 × 10^-6
= (1.3/25) 10^-5 × 10^6
= 0.052× 10
= 0.52m
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s
Answer:
a) [tex]v(2) = 3.9\,\frac{m}{s}[/tex], b) [tex]v(4) = -15.7\,\frac{m}{s}[/tex]
Explanation:
a) The equation for vertical velocity is obtained by deriving the function with respect to time:
[tex]v(t) = 23.5 -9.8\cdot t[/tex]
The velocities at given instants are, respectivelly:
[tex]v(2) = 3.9\,\frac{m}{s}[/tex]
[tex]v(4) = -15.7\,\frac{m}{s}[/tex]
g A particle starts moving from the origin of the coordinate system with the initial velocity v(0)=<0,0,2> and acceleration at time t given by a(t)=<1,2,0> find the moment of time t=T when the particle hits the plane 2x+y-z=4
Answer:
2s
Explanation:
The position function of the motion can be expressed as:
[tex]s = s_0 + v_0t + at^2/2[/tex]
where [tex]s_0 = <0,0,0>[/tex] is the origin where the particle starts off, [tex]v_0 = <0,0,2> m/s[/tex] and a = <1,2,0> m/s2. In the 3-coordinate system it can be written as
[tex]s = <0 + 0t+ t^2/2, 0 + 0t + 2t^2/2, 0 + 2t + 0t^2/2> = <t^2/2, t^2, 2t>[/tex]
For the particle to hit the 2x+y-z=4 plane then its coordinates must meet that criteria
[tex]2t^2/2 +t^2-2t = 4[/tex]
[tex]2t^2 - 2t -4 = 0[/tex]
[tex]t^2 - t - 2 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{1\pm \sqrt{(-1)^2 - 4*(1)*(-2)}}{2*(1)}[/tex]
[tex]t= \frac{1\pm3}{2}[/tex]
t = 2 or t = -1
Since t can only be positive we will pick t = 2s
Laminar flow of oil in a 2-in Schedule 40 steel pipe has an average velocity of 10.72 ft/s. Find the velocity at (a) the center of the pipe, (b) at the wall of the pipe, and (c) at a distance of 0.6 inches from the centerline.
Answer:
(a) 21.44 ft/s
(b) 0 ft/s
(c) 19.51 ft/s
Explanation:
2 in = 2/12 ft = 0.167 ft
For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:
[tex]v(r) = v_c\left[1 - \frac{r^2}{R^2}\right][/tex]
where R = 0.167 ft is the pipe radius and [tex]v_c[/tex] is the constant fluid velocity at the center of the pipe.
We can integrate this over the cross-section area of the in order to find the volume flow
[tex]\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\[/tex]
So the average velocity
[tex]v = \dot{V} / A = v_c/2 = 10.72[/tex]
[tex]v_c = 10.72*2 = 21.44 ft/s[/tex]
b) At the wall of the pipe, r = R so [tex]v(R) = v_c(1 - 1) = 0 ft/s[/tex]
c) At a distance of 0.6 in = 0.6/12 = 0.05 ft
[tex]v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s[/tex]
Answer:
The answers to the questions are;
(a) The velocity at the center of the pipe is 21.44 ft/s
(b) The velocity at the wall of the pipe is 0 ft/s
(c) The velocity at a distance of 0.6 inches from the center-line is 19.63 ft/s.
Explanation:
To solve the question, we note that
The velocity profile in the cross section of a circular pipe with laminar flow is given by
U = 2×v×[1 - (r/r₀)²]
Where
U = The sought velocity at a point
r = Pipe radius where velocity is sought
r₀ = Internal radius of pipe = for 2-in Schedule 40 steel pipe = 2.067 in 52.6 mm
v = Average velocity of flow = 10.72 ft/s = 3.2675 m/s
Therefore we have
(a) The velocity at the center of the pipe
At the center r = 0 so we have
U = 2×v×[1 - (r/r₀)²]
At center U = 2×10.72 ft/s×[1 - (0/2.067 in)²] = 2×10.72 ft/s = 21.44 ft/s
(b) The velocity at the wall of the pipe is given by
r = r₀ ⇒ U = 2×v×[1 - (r/r₀)²] ⇒ U = 2×v×[1 - (r₀/r₀)²]
= U = 2×v×[1 - (1)²] = 2×v×0 = 0
The velocity at the wall of the pipe is 0 ft/s
(c) The velocity at a distance of 0.6 inches from the center-line is given by
U = 2×v×[1 - (r/r₀)²] = 2×10.72 ft/s×[1 - (0.6/2.067)²] = 19.63 ft/s.