Answer:
The volume is 2.22x10⁵m³
Explanation:
the solution is in the attached Word file
To calculate the volume of space containing 1.0 J of electromagnetic energy near the Earth, use the formula Volume = Energy / Intensity after converting the intensity from kW/m² to W/m².
The volume of space that contains 1.0 J of electromagnetic energy can be found using the formula:
Volume = Energy / Intensity
Given that the intensity is 1.35 kW/m², you need to convert it to W/m² (1 kW = 1000 W).
After converting the intensity, you can then calculate the volume using the given energy of 1.0 J.
The volume of space near Earth containing 1.0 J of electromagnetic energy can be found by first calculating the energy density using the given intensity and speed of light, and then dividing the amount of energy by energy density, resulting in a volume of 2.22 imes 10^5 m^3.
Finding the Volume of Space Containing 1.0 J of Electromagnetic Energy
To find the volume of space near Earth that contains 1.0 J of electromagnetic energy given the intensity (I) from the sun as 1.35 kW/m2, we can use the formula for energy density (u), which is given by the equation u = I/c, where c is the speed of light. Now, to find the volume (V) that contains energy (E), we use the equation V = E/u.
First, let's convert the intensity to watts per square meter: 1.35 kW/m2 = 1350 W/m2. Next, calculate the energy density (u):
u = 1350 W/m2 / (3.0 * 108 m/s) = 4.5 * 10-6 J/m3.
With the energy density known, we can calculate the volume (V) that contains 1.0 J of energy:
V = 1.0 J / (4.5 * 10-6 J/m3) = 2.22 * 105 m3.
Therefore, a volume of 2.22 * 105 m3 near Earth contains 1.0 J of electromagnetic energy.
In empty space, which quantity is always larger for X-ray radiation than for a radio wave? In empty space, which quantity is always larger for X-ray radiation than for a radio wave? Speed. Wavelength. Amplitude. Frequency.
In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.
Frequency is the number of complete cycles (oscillations) of a wave that occur in one second. It is measured in Hertz (Hz), where 1 Hz means one cycle per second. X-ray radiation has a higher frequency than radio waves.
On the electromagnetic spectrum, X-rays have much higher frequencies compared to radio waves. X-rays have frequencies in the range of [tex]10^16[/tex] to [tex]10^19[/tex] Hz, while radio waves typically have frequencies in the range of [tex]10^3[/tex] to [tex]10^9[/tex] Hz. Therefore, X-ray radiation has significantly higher frequencies than radio waves.
Wavelength and amplitude are other properties of waves, but they are not always larger for X-ray radiation than for a radio wave. The wavelength of X-rays is much shorter than that of radio waves, and the amplitude can vary depending on the specific wave source and conditions.
Hence, In empty space, the quantity that is always larger for X-ray radiation than for a radio wave is the frequency.
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The sound level produced by one singer is 70.8 dB. What would be the sound level produced by a chorus of 37 such singers (all singing at the same intensity at approximately the same distance as the original singer)
Answer:
86.48 dB . Ans
Explanation:
Sound level by one singer = 70.8 dB
= 7.08 B
If I be its intensity in terms of watt/m²
log ( I / 10⁻¹² ) = 7.08 ( given )
I / 10⁻¹² = 10⁷°⁰⁸
I = 10⁻¹² x 10⁷°⁰⁸
= 10⁻⁴°⁹²
= .000012022 W / m²
intensity by 37 singers
= 37 x .000012022 W /m²
= .0004448378
= 4448.378 x 10⁻⁷
intensity level
= log I / 10⁻¹² B
= log 4448.378 x 10⁻⁷ / 10⁻¹²
= log 444837800 B
log 4448378 + 2 B
= 6.6482 +2 B
8.6482 B
intensity level = 8.6482 B
= 86.48 dB . Ans
Answer:
86.48 dB
Explanation:
86.48 dB
Explanation:
sound level intensity, β = 70.8 dB
Io = 10^-12 W/m²
Let the intensity if I at the level of 70.8 dB
Use the formula
7.08 = log I + 12
log I = - 4.92
I = 1.2 x 10^-5 W/m²
There are 37 singers, so the total intensity is I'.
I' = 37 x I
I' = 37 x 1.2 x 10^-5 = 4.45 x 10^-4 W/m²
Let the sound intensity is β' in decibels.
Use the formula
β' = 86.48 dB
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an angular displacement of 13 rad. What is its displacement θ at a time of t = 5.0 s?
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
Answer:
[tex]\theta = 52 rads[/tex]
Explanation:
The rotational kinetic equation is given by the formula:
[tex]\theta = w_{0} t + 0.5 \alpha t^{2}[/tex]..............(1)
The object is starting from rest, angular speed, w = 0 rad/s
At t = 2.5 sec, angular displacement, θ = 13 rad
Inserting these parameters into equation (1)
[tex]13 = (0 * 2.5) + 0.5 \alpha 2.5^{2}\\\alpha = \frac{13}{3.125} \\\alpha = 4.16 rad/s^{2}[/tex]
At time, t = 5.0 sec, we substitute the value of [tex]\alpha[/tex] into the kinematic equation in (1)
[tex]\theta = (0*5) + 0.5 *4.16* 5^{2}\\\theta = 52 rad[/tex]
An image is located at 4f, where f is the focal length, after it exits the lens. What is the object distance in terms of the focal length?
Answer:
p = 4/3 f
Explanation:
The constructor station describes the position of the image object in relation to the focal length
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image respectively
the objent distance is
1 / p = 1 / f - 1 / q
they indicate that the image is located at q = 4f this distance is positive because the image is after the lens
1 / p = 1 / f - 1 / 4f
1 / p = 3 / 4f
p = 4f / 3
the object is positive because it is to the left of the lens, according to the optical sign convention
The sign convention is that the distance to the objects is positive if it is on the left side of the lens and the image is positive if it is on the right side of the lens.
Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.
Answer:
Velocity between points A and B will be 0.2344 m/s
Complete Question:
A tennis racket is thrown vertically into the air. The center of gravity G has a velocity of 5 m/s upwards. Angular velocity about the x - direction of 1 rad/s and angular velocity about the y - direction of 20 rad/s. Knowing that the horizontal distance between points A and G as well as G and B is 25 cm and knowing that the maximum width of the of the racket head is 30 cm, determine the velocity of Points A and B.
Answer:
a) [tex]v_{A} = 10 m/s[/tex]
b)
[tex]v_{B} = 3i + 0.15 j m/s\\v_{B} = \sqrt{3^{2} + 0.15^{2} }\\v_{B} = 3.004 m/s[/tex]
Explanation:
a) Velocity of point A
The velocity of point A is a combination of the translational and the rotational velocity of the racket.
[tex]v_{A} = v_{G} + v_{r}[/tex]
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = wr[/tex]
r = 25 cm = 0.25 m
w = 20 rad/s
[tex]v_{r} = 20 * 0.25\\v_{r} = 5 m/s[/tex]
[tex]v_{A} = 5 + 5\\v_{A} = 10 m/s[/tex]
b) Velocity of point B
At point B, the linear velocity is in the +ve z-direction while the rotational velocity is in the -ve z-direction:
[tex]v_{G} = 5 m/s[/tex]
[tex]v_{r} = -r w\\v_{r} = - 0.25 * 20\\v_{r} = - 5 m/s[/tex]
[tex]v_{Bz} = v_{G} + v_{r} \\v_{Bz} = 5 -5\\v_{Bz} = 0 m/s[/tex]
In the y - direction, r = 30/2 = 15 cm = 0.15 m
r = 0.15 m
[tex]w_{x} = 1 rad/s[/tex]
[tex]v_{By} = rw_{x} \\v_{By} = 0.15 * 1\\v_{By} = 0.15 rad/s[/tex]
In the x - direction, r = 0.15 m, [tex]w_{y} = 20 rad/s[/tex]
[tex]v_{Bx} = rw_{y} \\v_{Bx} = 0.15 * 20\\v_{Bx} = 3.0 rad/s[/tex]
[tex]v_{B} = 3 i +0.15 j\\[/tex] m/s
A stainless steel tube with an outside diameter of 54 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 190 MPa, determine the maximum load P that the member can support.
Answer:146.26 kN
Explanation:
Given
Outside diameter of tube [tex]d_o=54\ mm[/tex]
thickness of tube [tex]t=5\ mm[/tex]
therefore inner diameter [tex]d_i=54-2\times 5[/tex]
[tex]d_i=44\ mm[/tex]
Cross-section of tube [tex]A=\frac{\pi }{4}(D_o^2-d_i^2)[/tex]
[tex]A=\frac{\pi }{4}\times 980[/tex]
[tex]A=245\pi mm^2[/tex]
Stress developed must be less than the limited value
thus [tex]\frac{P}{A}\leq \sigma [/tex]
[tex]P\leq \sigma A[/tex]
Maximum value of [tex]P=\sigma \times A[/tex]
[tex]P=190\times 245\times \pi [/tex]
[tex]P=146.26\ kN[/tex]
Final answer:
To calculate the maximum load a stainless steel tube can support, one must first find the cross-sectional area based on its outer diameter and wall thickness. Then, using the limit of the normal stress and the area, the maximum axial load can be computed.
Explanation:
To determine the maximum load P that a stainless steel tube can support when it is used as a compression member, we need to use the formula for axial stress in a cylindrical member, which is σ = P/A, where σ is the normal stress and A is the cross-sectional area. Given that the normal stress must not exceed 190 MPa, we need to first calculate the cross-sectional area. For a tube with an outside diameter (d) of 54 mm and a wall thickness (t) of 5 mm, the internal diameter (di) would be d - 2t = 54 mm - (2 x 5 mm) = 44 mm. The cross-sectional area A can be calculated using the formula for the area of a hollow circle, A = π/4 * (d2 - di2).
A = π/4 * ((55)^2 - (44)^2)
A = 855.298 * 10^-6 m2
Once we have the area, we can calculate the maximum load by rearranging the stress formula to P = σ * A. By plugging in the values for σ and A, we can find the value of P that the member can safely support. Remember to convert diameters to meters when calculating the area in square meters for consistency with the stress units of Pascals (Pa).
P = 190 * 855.298 Pa
P = 1.62 MPa
A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.8 cm away, in a time 3.5 10-7 s. What is the magnitude of the electric field
Answer:
3.53 N/C
Explanation:
Electric field = F / q where F is the force in N and q is charge on the electron
F = mass of an electron × a ( acceleration in m/s)
using equation of motion to solve for the acceleration
s ( distance ) = ut + 0.5 at² since the electron is starting from rest then ut = 0
2s / t² = a
F = me × ( 2s / t²)
E electric field = me × ( 2s / t²) / q = me × 2s / ( t² × q)
me, mass of an electron = 9.11 × 10⁻³¹ kg
E = (9.11 × 10⁻³¹ kg × 2 × 0.038 m) / ( (3.5 × 10⁻⁷s)² × 1.6 × 10⁻¹⁹ C) = 0.0353 × 10² N/C = 3.53 N/C
Answer: The magnitude of the electric field is 3.53 N/C
Explanation: Please see the attachments below
Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 m/s.They collide head-on and bounce apart elastically. By squeezing one of the balls in a vise while precise measurements are made of the resulting amount of compression, you find that Hooke's law is a good model of the ball's elastic behavior. A force of 15.9 kN exerted by each jaw of the vise reduces the diameter by 0.130 mm. Model the motion of each ball, while the balls are in contact, as one-half of a cycle of simple harmonic motion. Compute the time interval for which the balls are in contact.
Answer:
Explanation:
To detrmine the time interval at which the balls are in contact.
Given information
The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.
Expression for the effective spring constant ball is shown below.
K = |F|/|x|
Here,
k is a spring constant
F is the force exerted on the ball
x is dispalcement due force
substitute 15.9 kN for F and 0.130 mm in above equation
K = (15.9 kN)(1X10³N) / (0.130 mm)(1x10⁻³m/1mm)122 x 10⁶ N/mThe spring constant is 122 x 10⁶ N/m
A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each with a radius of 5.00 mm. If water flows through the single pipe at 1.45 m/s, calculate the speed (in m/s) of the water in the narrower pipes.
Answer:
1.24 m/s
Explanation:
Metric unit conversion:
9.25 mm = 0.00925 m
5 mm = 0.005 m
The volume rate that flow through the single pipe is
[tex]\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s[/tex]
This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of
[tex]\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s[/tex]
So the flow speed of each of the narrower pipe is:
[tex]v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}[/tex]
[tex]v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s[/tex]
A 0.74-m diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 m*N which accelerates it uniformly from rest through a total of 160 revolutions in 14.0 s.
1. What is the mass of the sphere?
Answer:
Mass of the sphere is 19.2 kg
Explanation:
We have given diameter of the sphere d = 0.74 m
So radius r = 0.37 m
Initial angular velocity [tex]\omega _i=0rad/sec[/tex]
Time t = 14 sec
Angular displacement [tex]\Theta =160revolution=160\times 2\pi =1004.8rad[/tex]
From second equation of motion
[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2[/tex]
So [tex]1004.8=0\times t+\frac{1}{2}\times \alpha \times 14^2[/tex]
[tex]\alpha =10.25rad/sec^2[/tex]
Torque is given [tex]\tau =10.8Nm[/tex]
Torque is equal to [tex]\tau =I\alpha[/tex], here I is moment of inertia and [tex]\alpha[/tex] angular acceleration
So [tex]10.8=10.25\times I[/tex]
[tex]I=1.053kgm^2[/tex]
Moment of inertia of sphere is equal to [tex]I=\frac{2}{5}Mr^2[/tex]
So [tex]1.053=\frac{2}{5}\times M\times 0.37^2[/tex]
M = 19.23 kg
So mass of the sphere is 19.23 kg
A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?
Answer:
What displacement must the physics professor give the car
= 12.91 METERS
Explanation:
Check the attached file for explanation
Final answer:
To find the displacement that the physics professor must give the car, we can break down the different stages of the car's motion and use the equations of motion. By adding the displacements from each stage, we can find the total displacement of the car.
Explanation:
To find the displacement that the physics professor must give the car in order for its average velocity to be +6.50 m/s for the entire trip, we can break down the different stages of the car's motion and use the equations of motion.
1. From time t=0 to t=3.60s: The car is accelerated in the +x direction at 5.35 m/s². We can use the equation v = u + at to find the final velocity v1 at t=3.60s, where u is the initial velocity, a is the acceleration, and t is the time.
2. From time t=3.60s to t=8.10s: The car slows down at a rate of 1.95 m/s² due to friction. We can use the equation v = u + at to find the initial velocity u2 at t=3.60s, where v is the final velocity, a is the acceleration, and t is the time.
3. From time t=8.10s to t=12.60s: The physics professor pushes the car with a constant velocity for 4.50s. The average velocity during this time is +6.50 m/s. We can calculate the displacement during this time using the equation d = vt, where v is the velocity and t is the time.
By adding the displacements from each stage, we can find the total displacement of the car, which is the answer to the question.
At some point, not close to its ends, within a solenoid of arbitrary length, calculate the approximate magnetic field if the solenoid carries a current 10.0 A and has 110.0 turns per meter of the solenoid's length.
Answer: The approximate magnetic field is 1.38 × 10^-3T
Explanation: Please see the attachment below
The approximate magnetic field at a point within the solenoid is 4.4 × 10^-4 T.
Explanation:To find the magnetic field strength at a point within a solenoid, we can use the equation B = µ0nI, where B is the magnetic field strength, µ0 is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid carries a current of 10.0 A and has 110.0 turns per meter of its length, we can substitute these values into the equation to calculate the magnetic field strength:
B = (4π × 10-7 T·m/A)(110.0 turns/m)(10.0 A) = 4.4 × 10-4 T
Therefore, the approximate magnetic field at a point within the solenoid is 4.4 × 10-4 T.
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A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present-day digital camera, the CCD chip) in order to focus an object that is infinitely far away (namely the incoming light rays are parallel with the principal axis of the system). (b) How far should the lens be from the film to focus an object at a distance of 2f in front of the lens
Answer:
a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.
b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.
Explanation:
To focus an infinitely distant object, the lens should be placed at its focal length from the film or CCD. For an object at 2f, the lens should also be at a distance of 2f to focus the image correctly.
The question pertains to the principles of optics, specifically focusing using converging lenses in cameras.
(a) To focus an object that is infinitely far away, the lens should be placed at its focal length f from the film or CCD. This is because light rays from a distant object enter the lens parallel to its principal axis and converge at the focal point on the opposite side. Therefore, for an object at infinity, the image distance (di) approaches the lens's focal length (f).
(b) To focus an object at a distance of 2f in front of the lens, the lens should also be 2f from the film. This is derived from the lens formula (1/f = 1/do + 1/di) where do is the object distance and di is the image distance. For an object at 2f, solving the equation gives di = 2f.
Gases are a unique state of matter because the volume of a gas can be compressed or expanded based on its temperature. Jacque Charles studied these changes by observing hot air balloons. For a given mass, as the gas is heated its volume expands and as it cools the volume contracts. Charles found that the volume is directly proportional to temperature when the mass and pressure remain constant. Which equation represents the relationship in Charles' law?
A)(P1V1/T1)=(P2V2/T2)
B) (V1/T1)=(V2/T2)
C) (V1/n1)=(V2/n2)
D) P1V1=P2V2
Answer:
B)
Explanation:
Hi there,
Notice in the prompt it is saying given mass, which means quantity of gas is constant. This eliminates option C, as this is Avogadro's Law (the more gas you have, the more space it takes up). It also states pressure remains constant, so it eliminates D, Boyle's Law (pressure is inversely related with volume), and A, The Combined Gas Law (combines Charles, Boyle's, and Gay-Lussac's Laws).
Charles' Law is telling us that as a gas is heated, the volume increases. This makes sense, as when gases are heated the average kinetic energy (energy of motion and movement) of the system is higher because of heat increase. When the system's kinetic energy is higher, it will have the tendency to occupy more space to balance out the increase in energy. Think of people who become more jittery in an already crowded space, they will push outward to accompany more space to balance.
So, answer is B.
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The number of tornado deaths in the United States in the 2000s was less than 40 percent the number that occurred in the 1950s, even though there was a significant increase in population. Suggest a likely reason for the decline in the death toll.
ALSO..
A television meteorologist is able to inform viewers about the intensity of an approaching hurricane. However, the meteorologist can report the intensity of a tornado only after it has occurred. Why is this the case?
Answer:
Answer
Explanation:
The followings are the reasons for the reduction in the death toll from Tornadoes occurence in the 2000s compared to 1950s;
(a) Improvement in technology and rapid natural occurence prediction. Through the use of advanced radar systems the early detection of natural occurence e.g tornadoes, providing warning signal to the residents and tracking their possible pathways is made possible.
Fast spread of information through television channels, Radio, Social media etc., help in fast spreading of information, thereby sending early signals to people thus making them prepare for its occurence.
Improvement of protective measures, provision of housing infrastructure for the displaced. With increase in technology, the protective measures are also improving.
Quick Government and medical interventions for casualties. Conducting Tornado drills help people to practice to take cover in a specified location during tornadoes.
TV meteorologist are able to inform viewers of the approaching tornadoes and hurricanes which can be tracked from the time of its origin i.e., from a depression to a storm. Advanced radar data help in delineating the possible pathways of the hurricane.
Tornadoes on the other hand form in a thunderstorm or during a frontal activity. In thunderstorms there are vortices formed which are pre-cursors of tornadoes.. A vortice when touches the ground forms a tornado. Once a tornado is on the ground it becomes easy for radars to measure windpeeds. Hence a TV meteorologist can report the intensity of the Tornado.
Difference Between Photometry and Spectrophotometry ?
Answer:
Spectrophotometry is the quantitative measurement of light spectra reflection and transmission properties of materials as function of the wavelength. Photometry measures the total brightness as seen by the human eye
Explanation:
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A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in contact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?
Answer:
1400 N
Explanation:
Change in momentum equals impulse which is a product of force and time
Change in momentum is given by m(v-u)
Equating this to impulse formula then
m(v-u)=Ft
Making F the subject of the formula then
[tex]F=\frac {m(v-u)}{t}[/tex]
Take upward direction as positive then downwards is negative
Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then
[tex]F=\frac {0.3(2--5)}{0.0015}=1400N[/tex]
The average force exerted on the ball by the floor is 1400 N.
What is force?Force is the product of mass and acceleration. The S.I unit of force is Newton (N)
To calculate the average magnitude of the force exerted on the ball by the floor, we use the formula below.
Formula:
F = m(v-u)/t............... Equation 1Where:
F = Force exerted on the ball by the floor.m = mass of the floorv = final velocity of the ballu = initial velocity of the ballt = timeFrom the question,
Given:
m = 0.3 kgv = 5 m/su = -2 m/s (rebounds)t = 1.5×10⁻³ sSubstitute these values into equation 1
F = 0.3[(5)-(-2)]/(1.5×10⁻³)F = 0.3(5+2)/(1.5×10⁻³)F = 2.1/(1.5×10⁻³)F = 1.4×10³F = 1400 NHence, the average force exerted on the ball by the floor is 1400 N.
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One can estimate the oil reservoir pressure underground from the height the oil rises. Of the gusher rises to 200 ft above ground and the diameter of the bore pipe that goes down to the oil reservoir underground is 9 inches, calculate the oil pressure in a reservoir that is 1500 m below ground. The oil density is 0.85 g/cm3 and viscosity is 200 cP
Answer:
12,608 kPa
Explanation:
First we need to convert the density to standard units, that is kg/m^3
0.85 g/cm^3 = 0.85 × [tex]\frac{100^{3} }{1000}[/tex] = 0.85 × 1000 = 850 kg/m^3
Pressure of oil = pressure at surface + rho × g × h
= 101,000 + (850×9.81×1500)
= 12,608 kPa
The units in the part 'rho.g.h' will cancel out against each other and you will be left with the unit Pascals - which is the unit for pressure.
Hope that answers the question, have a great day!
A plane electromagnetic wave varies sinusoidally at 86.0 MHz as it travels through vacuum along the positive x-direction. The peak value of the electric field is 2.30 mV/m, and it is directed along the positive y-direction. Determine the average power per unit area (the intensity) this wave carries through space. (Be careful with your units here and make sure to submit your answer in μW/cm2)
Answer:
The intensity is [tex]I = 0.0003053 \mu W/cm^2[/tex]
Explanation:
From the question we are told
The frequency of the electromagnetic wave is [tex]f = 86.0 Hz[/tex]
The peak value of the electric field is [tex]E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m[/tex]
Generally the intensity of this wave is mathematically represented as
[tex]I = c * \frac{1}{2} * \epsilon_o E^2_o[/tex]
Where c is the speed of light with value [tex]c = 3 *10^8 m/s[/tex]
[tex]\epsilon_o[/tex] is the permittivity of free space with value [tex]\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2[/tex]
Substituting values into equation for intensity
[tex]I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}[/tex]
[tex]I = 3.053 *10^{-6} W/m^2[/tex]
Converting to [tex]cm^2[/tex] we divide by 10,000
[tex]I = \frac{3.053 *10^{-6}}{10000} W/cm^2[/tex]
[tex]= 3.053 *10^{-10} W/cm^2[/tex]
[tex]= 0.0003053 *10^{-6} W/cm^2[/tex]
[tex]I = 0.0003053 \mu W/cm^2[/tex]
Suppose we assume that Einstein and Zweistein should compute the same total force on the sphere. Would Zweistein compute a larger or smaller value of the electric field at the sphere than Einstein
Answer:
Forces observed by both is same as both
The frames are non-acceleration
Explanation:
Zweistein = Einstein
[tex](F_B + E_B)_Z[/tex] = [tex](F_B + E_B)_E[/tex]
as [tex](F_B + E_B)_E[/tex] tend to zero (0)
[tex](E_B)_Z[/tex] < [tex](F_B)_Z[/tex] as some quantity is subtracted from the force.
According to Einstein's Theory of Relativity, the laws of physics are the same for all observers, regardless of their motion. Applying this to the calculation of electric field strength around a sphere, both Einstein and Zweistein, assuming they are not accelerating, should compute the same value of the electric field.
Explanation:If we consider the principles of physics consistently at play regardless of the observer, both Einstein and Zweistein must compute the same total force on the sphere. This is following Einstein's postulate stating the laws of physics remain the same in all inertial frames. Therefore, the electric field computed at the sphere by both Einstein and Zweistein should also be the same.
The principle of relativity implies that the laws of physics—including those of electric field strength—are the same for all observers, regardless of their speed or direction as long as they are not accelerating. Einstein's Theory of Relativity extends this to include accelerated motion and gravity.
This principle, combined with the equivalence of inertial and gravitational mass, is enough to show that the force of gravity perceived by an observer in an accelerating spacecraft must be the same as if the observer were stationary on the Earth. So, if Einsteins's Theory of Relativity applies to electric fields in the same way it does for gravitational fields, then the electric field strength calculated by Einstein and Zweistein should be the same.
To clarify, in the context of this question, the 'electric field' refers to the region around a charged particle or object where another charge placed within it would experience a force either of attraction or repulsion.
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A distance-time graph indicates that an object moves 20 m in 10 s and then remains at rest for 20s. What is the average speed of the object?
Answer:
v = 2 m/s
Explanation:
The object moves 20 m in 10 s initially. Then it remains at rest for 20 s. It is required to find the average speed of the object.
Fro 10 s it covers 20 m. For that time, speed is :
[tex]v=\dfrac{d}{t}\\\\v=\dfrac{20\ m}{10\ s}\\\\v=2\ m/s[/tex]
For 20 s, it was at rest means distance covered is 0. So, speed is 0.
The average speed of the object is equal to the sum of speed for 10 s and for 20 s i.e.
v = 2 m/s + 0 = 2 m/s
So, the average speed of the object is 2 m/s.
A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.Part A
The induced emf in the loop is measuredto be V. What is the magnitude B of the magnetic field that the loop was in?
Part B
For the case of a square loop of sidelength L being pulled out of the magneticfield with constant speed v, what is the rate of change ofarea c = -\frac{da}{dt}?
The question is not clear enough. So i have attached a copy of the correct question.
Answer:
A) B = V/c
B) c = Lv
Explanation:
A) we know that formula for magnetic flux is;
Φ = BA
Where B is magnetic field and A is area
Now,
Let's differentiate with B being a constant;
dΦ/dt = B•dA/dt
From faradays law, the EMF induced is given as;
E = -dΦ/dt
However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.
Thus, V = -dΦ/dt
Thus, we can now say that;
-V = B•dA/dt
Now from the question, we are told that dA/dt = - c
Thus;
-V = B•-c
So, V = Bc
Thus, B = V/c
B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:
Thus;
E =LvB or can be written as; V = LvB
Where;
V is EMF
L is length of bar
v is velocity
From the first solution, we saw that;
V = Bc
Thus, equating both of the equations, we have;
Bc = LvB
B will cancel out to give;
c = Lv
The magnitude of the magnetic field B is calculated using Faraday's Law and is equal to V/-c. For a square loop being pulled out from the field, the rate of change of area is proportional to the speed of withdrawal and the length of the side still in the field. Hence, c = Lv.
Explanation:Part A: The magnitude of the magnetic field B can be calculated using Faraday's Law of electromagnetic induction, which states that the induced voltage in a circuit is equal to the negative rate of change of magnetic flux. Mathematically, it is expressed as V = -dB/dt, where V is the induced emf, and dB/dt is the rate of change of magnetic field. Because the area is decreasing at a constant rate, the magnetic field B is V/-c.
Part B: The rate of change of the area, denoted as c = -da/dt, for a square loop of side length L being pulled out of the magnetic field with constant speed v, could be calculated using the formula c = Lv. This is because as the square loop is pulled out with a constant velocity, the rate at which its area decreases is proportional to its speed and the length of the side that is still in the field.
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A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a .. 29.47 particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) W
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.10 cm, and at a particular instant the conduction current in the wires is 0.276 A. (a) What is the displacement current density jD in the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis? (d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
Given Information:
Radius of parallel-plate capacitor = 4.10 cm = 0.041 m
Conduction current = Ic = 0.276 A
Required Information:
a) Displacement current density = JD = ?
b) Rate of change of electric field = dE/dt = ?
c) Magnetic field between plates at r = 2 cm = ?
d) Magnetic field between plates at r = 1 cm = ?
Answer:
a) Displacement current density = JD = 52.27 A/m²
b) Rate of change of electric field = dE/dt = 5.904×10¹² V/m.s
c) Magnetic field between plates at r = 2 cm = 6.568×10⁻⁷ Tesla
d) Magnetic field between plates at r = 1 cm = 3.284×10⁻⁷ Tesla
Explanation:
a) What is the displacement current density JD in the air space between the plates?
Displacement current density is given by
JD = Id/A
Where Id is the conduction current and A is the area of capacitor given by
A = πr²
A = π(0.041)²
A = 0.00528 m²
As you can notice in the diagram, conduction current has equal displacement between the capacitor plates therefore, Id = Ic
JD = 0.276/0.00528
JD = 52.27 A/m²
b) What is the rate at which the electric field between the plates is changing?
The rate of change of electric field is given by
dE/dt = JD/ε₀
Where JD is the displacement current density and ε₀ is the permittivity of free space and its value is 8.854×10⁻¹² C²/N.m²
dE/dt = 52.27/8.854×10⁻¹²
dE/dt = 5.904×10¹² V/m.s
c) What is the induced magnetic field between the plates at a distance of 2 cm from the axis?
The induced magnetic field between the plates can be found using Ampere's law
B = (μ₀/2)*JD*r
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷ T.m/A, JD is the displacement current density and r is the distance from the axis.
B = (4π×10⁻⁷/2)*52.27*0.02
B = 6.568×10⁻⁷ Tesla
d) What is the induced magnetic field between the plates at a distance of 1 cm from the axis?
B = (μ₀/2)*JD*r
B = (4π×10⁻⁷/2)*52.27*0.01
B = 3.284×10⁻⁷ Tesla
This physics-based question involves understanding the properties and behaviors of a parallel-plate, air-filled capacitor. Specifically, it focuses on finding the displacement current density and the rate at which the electric field between the plates is changing as the capacitor is charged.
Explanation:The question revolves around a parallel-plate capacitor, a system of two identical conducting plates separated by a distance. This specific case involves an air-filled capacitor where the plates have a certain radius, and a conduction current is defined.
First, let's address part (a): What is the displacement current density jD in the air space between the plates? The displacement current density, noted as 'jD', can be found by using Maxwell's equation which correlates the time rate of change of electric flux through a surface to the displacement current passing through the same surface.
For (b): What is the rate at which the electric field between the plates is changing? As the current is introducing charge onto the plates, the electric field (which is proportional to the charge on the capacitor's plates) will be changing. This change can be calculated as the derivative of the electric field's magnitude with respect to time.
Without specific numerical values or more information, I'm unable to provide a more detailed analysis or a concrete answer.
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An object moving with a speed of 21 m/s and has a kinetic energy of 140 J, what is the mass of the object.
Answer:2940
Explanation:
Answer:
0.635 kg
Explanation:
Recall that Kinetic energy is defined as:
K.E. = (1/2) mv²,
where
K.E = Kinetic energy = given as 140J
m = mass (we are asked to find this)
v = velocity = given as 21 m/s
Substituting the above values into the equation:
K.E. = (1/2) mv²
140 = (1/2) m (21)² (rearranging)
m = (140)(2) / (21)²
m = 0.635 kg
The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts horizontally on an object on the Moon, producing an acceleration equal to am. In the absence of friction and drag, the same force acting on the same object on Earth would produce an acceleration i.e. equal to
a. 6am
b. (1/6)am
c. am
d. 0
e. 9.8 m/s2
Answer:
c.[tex]a_m[/tex]
Explanation:
We are given that
Acceleration due to gravity on the moon=[tex]a_m[/tex]
Acceleration due to gravity on the earth=[tex]a_e[/tex]
[tex]g_m=\frac{1}{6}g_e[/tex]
Net force due to am on an object on moon=[tex]F_{net}=ma_m[/tex]
There is no friction and no drag force and there is no gravity involved
Then, the force acting on an object on earth=[tex]F=ma_e[/tex]
[tex]F=F_{net}[/tex](given)
[tex]ma_m=ma_e[/tex]
[tex]a_e=a_m[/tex]
Hence, option c is true.
A 10.0 μFμF capacitor initially charged to 30.0 μCμC is discharged through a 1.30 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 15.0 μCμC ? Express your answer with the appropriate units. 0.052s
Answer:
It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.
Explanation:
Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit ([tex]\tau_0=RC[/tex]) which in our case is;
[tex]\tau_0=1.3\,k\Omega\,*\,10.0 \mu F=13\,\,10^{-3}\,s[/tex] (13 milliseconds)
Recall as well the expression for the charge in the capacitor (from it initial value [tex]Q_0[/tex], as it discharges via a resistor R:
[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }[/tex]
So knowing that the capacitor started with a charge of 30 [tex]\mu C[/tex] and reduced after a time "t" to 30 [tex]\mu C[/tex] , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:
[tex]Q(t)=Q_0\,e^{-\frac{t}{RC} }\\15 \mu C=30 \mu C\,\,e^{-\frac{t}{13\,ms} }\\e^{-\frac{t}{13\,ms} }=\frac{1}{2} \\-\frac{t}{13\,ms}=ln(\frac{1}{2})\\t=-13\,ms\,*\,ln(\frac{1}{2})\\t=9.109\,\, ms[/tex]
In seconds this is approximately 0.009 seconds
Final answer:
The question involves calculating the time it takes for the charge on a capacitor to reduce to a certain value in an RC circuit, using the natural logarithm and the RC time constant formula. This is a Physics problem relevant to High School level.
Explanation:
The subject matter of the student's question is Physics, specifically relating to the discharge of a capacitor through a resistor and the calculation of time constants in RC circuits. The grade level would be High School, as it involves concepts typically taught in a high school physics course.
How to Calculate the Discharge Time
To calculate the time it takes for a capacitor to reduce its charge to a certain value, we use the formula V = V0 * e^(-t/RC), where V0 is the initial voltage, V is the final voltage, R is the resistance, C is the capacitance, t is the time, and e is the base of the natural logarithm. However, since we're dealing with the capacitor's charge rather than voltage, we can adapt the formula to Q = Q0 * e^(-t/RC). By substituting the given values and solving for t, we can find the time it takes for the capacitor's charge to decrease from 30.0 μC to 15.0 μC.
To solve for t, we would take the natural logarithm of both sides after dividing by Q0, resulting in t = -RC * ln(Q/Q0). Inserting the given values (R = 1.30 kΩ and C = 10.0 μF), and calculating the natural logarithm of (Q/Q0 = 15.0 μC / 30.0 μC), we find the time required for the charge to drop to half its initial value.
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole’s symmetry axis. The rope makes a constant angle of 70.0° with respect to the vertical. What is the tangential speed of the ball?
Answer:
The tangential speed of the ball is 11.213 m/s
Explanation:
The radius is equal:
[tex]r=2.3*sin70=2.161m[/tex] (ball rotates in a circle)
If the system is in equilibrium, the tension is:
[tex]Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}[/tex]
Replacing:
[tex]\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}[/tex]
Replacing:
[tex]v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s[/tex]
An object with a mass of 2.3 kg has a force of 6.2 newtons applied to it. What is the resulting acceleration of the object?
Answer:2.7m/s^2
Explanation:
mass=2.3kg
Force=6.2Newton
Acceleration=force ➗ mass
Acceleration=6.2 ➗ 2.3
Acceleration=2.7m/s^2
The resulting acceleration of the object is 2.70 m/s^2.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given that:
Mass of the object: m = 2.3 kg.
Force applied on it: F = 6.2 Newtons.
Now from Newton's 2nd law of motion: it can be stated that:
force = mass × acceleration
acceleration = force/mass
= 6.2 Newton/2.3 kg
= 2.70 m/s^2.
Hence, the acceleration of the object is 2.70 m/s^2.
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When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by n , gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light c in vacuum to the speed v in the material, orn=cv
What is the minimum value that the index of refraction can have?
Answer:
First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c). The value of the refractive index of the medium is a measure of its "optical density": Light spreads at maximum speed in a vacuum but slower in others transparent media; therefore in all of them n> 1. Examples of typical values of are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).
The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:
The limit or minimum angle, α lim, is defined as the angle of refraction from which the refracted ray disappears and all the light is reflected. As in the maximum value of angle of refraction, from which everything is reflected, is βmax = 90º, we can know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:
βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1
Explanation:
When a light ray strikes the separation surface between two media different, the incident beam is divided into three: the most intense penetrates the second half forming the refracted ray, another is reflected on the surface and the third is breaks down into numerous weak beams emerging from the point of incidence in all directions, forming a set of stray light beams.
The minimum value of the index of refraction is 1, which occurs in a vacuum where the speed of light is at its maximum and not slowed by any material.
The minimum value that the index of refraction can have is for a vacuum, where light travels at its fastest and the speed of light is not slowed down by any medium. According to the definition of the index of refraction, n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material. Since the speed of light in a vacuum is the highest possible speed for light and cannot be exceeded, the index of refraction n has a minimum value of 1 in a vacuum. That is because when we plug in the values into the equation, with both c and v being equal (since v is the speed of light in a vacuum), the ratio n = c/v equals 1. Therefore, no material can have an index of refraction less than 1.
A thin metal cylinder of length L and radius R1is coaxial with a thin metal cylinder of length L and a larger radius R2. The space between the two coaxial cylinders is filled with a material that has resistivity rho. The two cylinders are connected to the terminals of a battery with potential difference ΔV, causing current I to flow radially from the inner cylinder to the outer cylinder. Part A Find an expression for the resistance of this device in terms of its dimensions and the resistivity. Express your answer in terms of some or all of the variables R1, R2, L, and rho. R = nothing
Answer:
The expression for resistance is [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
Explanation:
Generally flow of charge at that point is mathematically given as
[tex]J = \frac{I}{2 \pi r L}[/tex]
Where L is length of the cylinder as given the question
The potential difference that is between the cylinders is
[tex]\delta V = -E dr[/tex]
Where is the radius
Where E is the electric field that would be experienced at that point which is mathematically represented as
[tex]E = \rho J[/tex]
Where is the [tex]\rho[/tex] is the resistivity as given the question
considering the formula for potential difference we have
[tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]
To get V we integrate both sides
[tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]
[tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
According to Ohm law
[tex]V= IR[/tex]
Now making R the subject we have
[tex]R = \frac{V}{I}[/tex]
Substituting for V
[tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
The resistance of a coaxial cylindrical configuration with radii R1 and R2, length L, and resistivity rho, is given by the formula R = (rho / (2 * pi * ln(R2/R1))).
The question asks for an expression for the resistance of a coaxial cylindrical configuration, where the space between two metal cylinders of length L and radii R1 and R2 is filled with a material of resistivity rho. To find this, we utilize the formula for the resistance R of a material, which is R = rho * (L/A), where A is the cross-sectional area. However, since the current flows radially through the material between the cylinders, we need to consider the formula for resistance in terms of the radii and length of the cylinders. The resistance can be expressed as R = (rho * L) / (2 * pi * L * ln(R2/R1)), simplifying to R = (rho / (2 * pi * ln(R2/R1))).