B. The Earth radiates an amount of energy into space equal to the amount it receives.
Part of the solar energy is reflected by the Earth into space, this is known as albedo. The other part of the energy radiated by the Earth in the form of infrared radiation, is absorbed by the greenhouse gases, which cause most of this infrared radiation to be emitted into space. Therefore, the net flow of energy is zero.
An automobile has a vertical antenna 1.45 m long. The automobile travels at 90 km/h on a horizontal road where the Earth's magnetic field is 59.0 µT directed to the north and downward at an angle of 65.5° below the horizontal. Calculate the maximum magnitude of the induced emf in the antenna.
Answer:
[tex]9.05\times 10^{-4} V[/tex]
Explanation:
We are given that
Length of antenna,l=1.45 m
Earth's magnetic field,[tex]B=59\mu T=59\times 10^{-6} T[/tex]
[tex]1\mu=10^{-6}[/tex]
[tex]\theta=65.5^{\circ}[/tex]
Speed,v=90 km/h=[tex]90\times \frac{5}{18}=25m/s[/tex]
[tex]1km/h=\frac{5}{18}m/s[/tex]
Magnitude of maximum induced emf in the antenna=[tex]E=Blvcos\theta[/tex]
Substitute the values
[tex]E=59\times 10^{-6}\times 1.45\times 25cos65[/tex]
[tex]E=9.05\times 10^{-4} V[/tex]
The top surface of the passenger car of a train moving at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top surface is absorbing solar radiation at a rate of 200 W/m2, and the temperature of the ambient air is 308C. Assuming the roof of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface of the car.
Answer:
33.1 degrees celcius
Explanation:
Look up attached file
The equilibrium temperature of the train’s top cannot be calculated without the specific heat transfer due to convection. The Stefan-Boltzmann law allows us to calculate heat transfer from radiation but not convection.
Explanation:The key to solving this problem is the Stefan-Boltzmann law of radiation, which describes the rate of heat transfer by emitted radiation. In brief, P = σeAT⁴, where σ = 5.67 × 10⁻⁸ J/s.m².K⁴ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, and T is its temperature in Kelvins.
Given that the top surface of the car absorbs solar radiation at a rate of 200 W/m² and its dimensions, we can calculate the area of the surface. The surface, if deemed perfectly insulated, only loses heat through convection, not radiation. However, the amount of heat transfer due to convection is not provided in the question, making it impossible to calculate the exact equilibrium temperature of the top surface of the car.
The net rate of heat transfer takes into account both the temperature of the object and the temperature of its surroundings. If T₁ represents the temperature of the surface to be investigated, and T₂ the ambient temperature, without the value of heat lost through convection, we cannot calculate T₁. Therefore, more information would be required to answer your question fully.
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According to the theory, the average force Favg, applied to the mass multiplied by the duration the force is applied should equal the mass change in momentum. For each run calculate the change in momentum of the mass using the average force, then find the % difference between that and change in momentum using the change in velocity.
Answer:
The change in momentum for average force is: Δ[tex]P_{average force} = F_{avg}[/tex]Δt and it is calculated below.
For thick and thin spring, the %difference between change in momentum for average force and change in momentum for velocities is; 178.27% and 159.72% respectively
Explanation:
The average force applied to an object can be calculated by multiplying the average net external force by the time interval over which the force acts. The change in momentum can be calculated using the average force, as well as the change in velocity. The percentage difference between the two values can be found by dividing the difference by the change in momentum using the change in velocity and then multiplying by 100.
Explanation:The average force applied to an object can be calculated using the average net external force multiplied by the time interval over which the force acts. This relationship is known as the impulse-momentum theorem, which states that the change in momentum is equal to the impulse. Impulse is the product of the average force and the time interval.
To calculate the change in momentum of an object using the average force, you can multiply the average force by the duration the force is applied. This will give you the change in momentum caused by the force. You can then calculate the change in momentum using the change in velocity by multiplying the mass of the object by the change in velocity. Finally, you can find the percentage difference between the two values by dividing the difference between the two values by the change in momentum using the change in velocity and then multiplying by 100.
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A leaf spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 6.2 cm (0.062 m) from its original unstretched position. What is the increase in potential energy of the spring (in kJ)?
The increase in potential energy of the spring when compressed a distance of 0.062m from its unstretched position, with a spring constant of 87600N/m, is approximately 0.168 kJ.
Explanation:The increase in potential energy of the spring can be calculated using the formula U = ½kx². In this problem, k is the spring constant which is 87,600 N/m and x is the displacement of the leaf spring when it is compressed, which is 0.062 m. Plugging in these values, we get U = ½ * (87,600 N/m) * (0.062 m)² = 167.5424 Joules. This value is in Joules, so to convert it to kilojoules (kJ), we divide by 1000. So, the increase in potential energy of the spring is approximately 0.168 kJ.
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The magnetic flux through a loop:
A. is maximized when the B field lines in the plane of the loop
B. is zero when the B field is perpendicular to the plane of the loop
C. doesn't depend on the direction of the B field
D. depends just on the B field going through the loop
E. is proportional to the number of magnetic field lines going through it
Answer:
The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.
Explanation:
Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.
It is expressed mathematically as
Φ = B A cos(θ) where
Φ is the magnetic flux
B is the magnetic field strength
A is the area
θ is the angle that the magnetic field make with the plane of the loop
If B is acting perpendicular to the plane of the loop, this means that θ = 90°
Magnetic flux Φ = BA cos90°
Since cos90° = 0
Φ = BA ×0
Φ = 0
This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.
Answer:
D. depends just on the B field going through the loop
Explanation:
The magnetic flux trough a loop is given by the formula:
[tex]\Phi_B=\vec{B}\cdot \vec{A}[/tex]
[tex]\Phi_B=BAcos\theta[/tex]
Where B is the magnitude of the magnetic field and A is the area of the loop.
It is clear that the flux is maximum when the angle between the direction of B and the direction of the normal vector of A is zero (cos0 = 1).
Furthermore, we can notice that only the magnetic field that crosses the loop contributes to the flux.
Hence, the correct answer is:
D. depends just on the B field going through the loop
hope this helps!!
A car is moving down a flat, horizontal highway at a constant speed of 21 m/s when suddenly a rock dropped from rest straight down from a bridge crashes through the windshield. The rock strikes the windshield at a right angle, and the windshield makes a 60 degree angle with the horizontal.
a. Find the speed of the rock relative to the car when it hits the windshield.
b. Find the distance above the point of contact with the windshield from which the rock was dropped.
c. If the driver had seen what was about to happen just as the rock was being released, how much time would she have had to react before the rock strikes the windshield?
Answer:
a)9.8
Explanation:
because any object falling from any height is 9.8
(sorry that's the only one i know)
1. The speed of the rock relative to the car when it hits the windshield is [tex]\( {42.00 \, \text{m/s}} \).[/tex]
2. The distance above the point of contact with the windshield from which the rock was dropped is [tex]\( {67.5 \, \text{m}} \).[/tex]
3. The driver would have had [tex]\( {3.71 \, \text{s}} \)[/tex] to react before the rock strikes the windshield.
Let's solve each part of the problem step-by-step.
Part (a): Find the speed of the rock relative to the car when it hits the windshield.
Given:
- Car speed, [tex]\( v_{\text{car}} = 21 \, \text{m/s} \)[/tex]
- Windshield angle with the horizontal,[tex]\( \theta = 60^\circ \)[/tex]
Since the rock is dropped from rest, its initial vertical velocity is [tex]\( 0 \, \text{m/s} \).[/tex]
The rock falls under the influence of gravity, so its vertical velocity [tex]\( v_{\text{vertical}} \)[/tex] at any time t is given by:
[tex]\[ v_{\text{vertical}} = g t \][/tex]
where [tex]\( g = 9.8 \, \text{m/s}^2 \).[/tex]
The horizontal velocity of the rock is [tex]\( 0 \, \text{m/s} \)[/tex] with respect to the ground.
To find the speed of the rock relative to the car when it hits the windshield, we need to consider the components of velocity with respect to the car:
1. The car's horizontal velocity component.
2. The rock's vertical velocity component.
The velocity components with respect to the car:
- Horizontal component relative to the car, [tex]\( v_{\text{horizontal, relative}} = v_{\text{car}} \)[/tex]
- Vertical component, [tex]\( v_{\text{vertical}} = g t \)[/tex]
The speed of the rock relative to the car can be found using the Pythagorean theorem:
[tex]\[ v_{\text{relative}} = \sqrt{(v_{\text{horizontal, relative}})^2 + (v_{\text{vertical}})^2} \][/tex]
We need to find t when the rock strikes the windshield.
Since the rock strikes the windshield at a right angle, and the windshield makes a 60° angle with the horizontal, the relative speed of the rock [tex]\( v_{\text{relative}} \)[/tex] must have components that align with the 60° angle.
Thus:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{relative}} \sin(60^\circ) \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{relative}} \cos(60^\circ) \][/tex]
Given:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{vertical}} = g t \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{car}} = 21 \, \text{m/s} \][/tex]
From the component relationship:
[tex]\[ \frac{v_{\text{relative, vertical}}}{v_{\text{relative, horizontal}}} = \tan(60^\circ) = \sqrt{3} \][/tex]
[tex]\[ \frac{g t}{21} = \sqrt{3} \][/tex]
[tex]\[ g t = 21 \sqrt{3} \][/tex]
[tex]\[ t = \frac{21 \sqrt{3}}{9.8} \][/tex]
Calculate t:
[tex]\[ t \approx \frac{21 \cdot 1.732}{9.8} \approx \frac{36.372}{9.8} \approx 3.71 \, \text{s} \][/tex]
Now, we can find the vertical velocity [tex]\( v_{\text{vertical}} \)[/tex]:
[tex]\[ v_{\text{vertical}} = g t = 9.8 \times 3.71 \approx 36.36 \, \text{m/s} \][/tex]
Now, the speed of the rock relative to the car:
[tex]\[ v_{\text{relative}} = \sqrt{(21)^2 + (36.36)^2} \approx \sqrt{441 + 1322.49} \approx \sqrt{1763.49} \approx 42.00 \, \text{m/s} \][/tex]
Part (b): Find the distance above the point of contact with the windshield from which the rock was dropped.
The distance h from which the rock was dropped can be found using the kinematic equation:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
Using [tex]\( t \approx 3.71 \, \text{s} \)[/tex]:
[tex]\[ h = \frac{1}{2} \times 9.8 \times (3.71)^2 \approx \frac{1}{2} \times 9.8 \times 13.76 \approx 67.5 \, \text{m} \][/tex]
Part (c): The time available for the driver to react is the time \( t \) it takes for the rock to fall, which we calculated in part (a):
[tex]\[ t \approx 3.71 \, \text{s} \][/tex]
(hrw8c13p51) A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as 0.11. What is the semimajor axis of the comet's orbit
Answer:
Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m
Explanation:
From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month
Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds
Now, the square of the period of any planet is given by the formula;
T² = (4π²/GM)α³
Where;
G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²
M is mass of sun which has a value of 1.99 x 10^(30) kg
α is the semi major axis of the comets orbit.
Thus, making α the subject, we have;
α = ∛(GMT²/4π²)
So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)
α = 1.89 x 10^(13) m
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Answer:
The order is 2>4>3>1 (TE)
Explanation:
Look up attached file
HELP ASAP PLEASE!!!
The earth is:
A. a perfect sphere
B. thinner at the equator
C. slightly flatter at the poles
D. made of mostly carbon and iron
The Earth is slightly flatter at the poles, making the correct answer option C. It is an oblate spheroid with a denser core made primarily of iron and nickel.
The correct answer to the question, 'The earth is:', is C: slightly flatter at the poles. This is because the Earth's shape is an oblate spheroid, meaning it's somewhat squished at the poles and bulges a bit at the equator due to its rotation. The Earth's equatorial radius is greater than its polar radius, which results in this slight flattening at the poles. Also, the Earth's core is composed predominantly of iron and nickel, making it very dense.
While it is true that the Earth is not a perfect sphere, it is also not thinner at the equator as option B suggests. Option D mentions that Earth is made of mostly carbon and iron, which is incorrect; the Earth is composed primarily of iron, oxygen, silicon, and other elements, but not mostly carbon.
4-True-False (experiment 6) : a) 1/(f^2) vs output current is a straight line ( f is frequency) b) Magnetic field of Helmholtz coil and earth are constant during the experiment. c) The sign of B_axial in experiment 6.2 can change the results for the value of N. d) In an ideal selenoid, the sign of B_radial is independent of sign of B_axial e) In experiment 6.1, If we use a magnetic dipole with parameters 4m and 2I, the intercept of 1/(T^2) vs current stays the same since the magnetic field of earth is constant.
Answer:
A.True
B. True
C False
D. False
E. True
Explanation:
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 46.8 H inductor, determine the inductive reactance and the rms current through the inductor.(a) the inductive reactance (in Ω)
(b) the rms current through the inductor (in A)
Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A
[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the staring for t seconds. Determine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.
Answer:
t = 5.89 s
Explanation:
To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.
Let us assume that the radius of the pulley ([tex]r_p[/tex]) = 0.4 m
Let the radius of the sphere (r) = 0.5 m
w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s
Tension (T) = 20 N
mass (m) = 3 kg each
[tex]\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw[/tex]
[tex]t = \frac{4r^2mw}{Tr_P}[/tex]
Substituting values:
[tex]t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s[/tex]
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value of t is [tex]t = 15.08 \ s[/tex]
Explanation:
From the question we are told that the angular speed is
The initial angular speed [tex]w_f = 150 rev/min = \frac{2 \pi }{60s} * 150 = 15.71rad /s[/tex]
The force is [tex]T = 20 N[/tex]
The radius is [tex]r = 400mm = \frac{400}{1000} = 0.4m[/tex]
The total mass of the four sphere is [tex]m_a = (4*3) = 12kg[/tex]
The initial velocity is [tex]v_i = 0m/s[/tex]
The radius of the pully is [tex]r_p = 100mm = \frac{100}{1000} = 0.10m[/tex]
The initial time is [tex]t_1 = 0s[/tex]
The final time is [tex]t = t[/tex]
Generally the final velocity of the sphere is mathematically represented as
[tex]v_f = r w_f[/tex]
[tex]v_f=15.7 r[/tex]
The angular impulse momentum principle can be represented methematically as
[tex](H_O)_i + \int\limits^{t_1}_{t_2}{(T \cdot r_{p})} \, dt = (H_O)_f[/tex]
[tex]r(m_a v_i ) + \int\limits^{t_2}_{t_1} {T \cdot r_p} \, dt = r(m_a v_f)[/tex]
[tex]r(m_a v_1 ) + T \cdot r_p (t_{2} -t_{1}) = r( m_a * 15.7 r )[/tex]
Substituting values
[tex]0.4 * (12* 0) + (20 *0.10 * (t-0)) = 0.40 * (12* 0.40 *15.7)[/tex]
=> [tex]0 + 2t = 30.16[/tex]
=> [tex]t = 15.08 \ s[/tex]
An earthquake on the ocean floor in the Gulf of Alaska induces a tsunami (sometimes called a "tidal wave") that reaches Hilo, Hawaii, 4780 km away, after a time of 9.99 h. Tsunami’s have enormous wavelengths (100- 200 km), and for such waves the propagation speed is v ≈ qgd , where d is the average depth of the water, d is in meters, v is in m/s, and g = 9.8 m/s^2 . From the information given, find the average ocean depth between Alaska and Hawaii.
Answer:
23785m
Explanation:
V= 4780/9.9= 482.8km/h
d= v²/g = 482.8²/9.8= 23785m
A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 77.9% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?
Answer:
(A) [tex]9.14\times 10^{-9}sec[/tex]
(B) [tex]6.20\times 10^{-3}A[/tex]
Explanation:
We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]
Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]
Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]
[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]
Battery voltage e = 16 volt
(a) It is given current becomes 79.9% of its final value
Current in RL circuit is given by
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
According to question
[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]e^{\frac{-t}{\tau }}=0.201[/tex]
[tex]{\frac{-t}{\tau }}=ln0.201[/tex]
[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]
[tex]t=9.14\times 10^{-9}sec[/tex]
(b) Current at [tex]t=\tau sec[/tex]
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]
[tex]i=6.20\times 10^{-3}A[/tex]
A 47 g particle undergoes SHM with an amplitude of 7.1 mm, a maximum acceleration of magnitude 5.4 x 103 m/s2, and an unknown phase constant φ. What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?
Answer:
Explanation:
check the picture attached for the explanation. I hope it helps . Thank you
Answer:
Explanation:
a) the period of the motion:
In order to calculate, we have the following formula.
f = 1/T whereas, T = 1/f. So, we need frequency first to calculate period of the motion.
In the question, we have been given amplitude and maximum acceleration. So, we can use the formula of maximum acceleration to calculate the frequency. Here's how:
Maximum acceleration = (2πf)² x A
where, A is amplitude.
a(max) = maximum acceleration
By rearranging the equation and making the frequency our subject and plugging in the values of given quantities then we have:
f = 138.80 hertz
Now, we can calculate, period of the motion by plugging in the value of frequency in the equation of period mentioned above.
Period of the motion = T = 1/f
T= 1/138.80
T = 7.204 x [tex]10^{-3}[/tex] seconds.
b) Maximum Speed of the particle
In order to calculate max. speed of the particle, we have to use following formula:
Max. Speed = 2πf x A
Max. Speed = ( 2 x 3.14 x 138.80 x 7.1x[tex]10^{-3}[/tex])
Max. Speed = 6.19 m/s
c) Total mechanical energy of the oscillator
Total mechanical energy of the oscillator is the sum of kinetic energy and potential energy and for which we have formula to calculate total mechanical energy of the oscillator:
Total Mechanical Energy = T.E = K.E + P.E
T.E = 2m[tex]\pi ^{2}[/tex][tex]f^{2}[/tex][tex]A^{2}[/tex]
we do have values of all the quantities and now by just plugging in the values we will get the total mechanical energy of the oscillator.
T.E = 2 x 0.047 x [tex]3.14^{2}[/tex] x [tex]138.80^{2}[/tex] x [tex]0.0071^{2}[/tex]
T.E = 0.90 J
d) Magnitude of the force on the particle when the particle is at its maximum displacement.
F = -KA here, amplitude is used because it is the maximum displacement from the mean position.
where K = -m ω² x A
ω = 2πf = 2 x 3.14 x 138.80
ω = 871.66
F = -(0.047 x 871.66² x 0.0071²)
F = - 1.8 N
e) when the displacement is at half
F = -Kx A/2
A/2 = 0.0071 / 2
A/2 = 0.0035
F = - m ω² x A/2
F = - (0.047 x 871.66² x 0.0035)
F = - 125 N
Assume that a uniform magnetic field is directed into this page. If an electron is released with an initial velocity directed from the bottom edge to the top edge of the page, which of the following describes the direction of the resultant force acting on the electron?
a. out of the page
b. to the right
c. to the left
d. into the page.
Final answer:
The direction of the resultant force acting on an electron moving upward on the page through a magnetic field directed into the page is to the left, as per the left-hand rule.
Explanation:
The question deals with the Lorentz force experienced by a charged particle, in this case an electron, moving through a magnetic field. When an electron with initial velocity directed from the bottom edge to the top edge of the page moves through a uniform magnetic field directed into the page, the force acting on it can be determined using the left-hand rule, since electrons have a negative charge. Holding your left thumb up (direction of the electron's velocity) and the fingers into the page (direction of the magnetic field), the resultant force will be directed to the left of the page. Therefore, the correct answer to the direction of the resultant force acting on the electron is c. to the left.
A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The battery has been connected for a long time. Each plate has an area A=58.3 m2 . The left plate is at ground potential and the right plate is at positive potential. The separation between the capacitor plates is D= 0.3m and all the points in the picture are far from the edges of the plates. The capacitor is placed on board of a space station where there is no gravity present.
In the coordinate system shown in the picture (x is a "horizontal" axis, y is "vertical" although names are purely conventional since there's no gravity), where all lengths are in meters, a positive charge q=3e-05C and mass m=0.4g has been placed directly in point C(0.25,12) and let go. What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
You decided to repeat the experiment but this time you place the same charge in point C first, and then connect the same power supply to the capacitor via the same resistor to charge it. Do not overcomplicated things: assume that the effect of any magnetic fields is negligible. Assuming that the axes scale on all graphs are the same what is the graph that best represents the horizontal speed of your charge as function of time in this experiment?
To verify your qualitative reasoning, calculate the speed of your charge in your second experiment after the same interval of time it took your charge to move from point C to point A in the first experiment. This is quite a challenging question so don't get disappointed if your group can't get the correct answer. You will not loose any points if you miss this question\
Answer:
See explanation
Explanation:
Given:-
- The DC power supply, Vo = 600 V
- The resistor, R = 1845 MΩ
- The plate area, A = 58.3 cm^2
- Left plate , ground, V = 0
- The right plate, positive potential.
- The distance between the two plates, D = 0.3 m
- The mass of the charge, m = 0.4 g
- The charge, q = 3*10^-5 C
- The point C = ( 0.25 , 12 )
- The point A = ( 0.05 , 12 )
Find:-
What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
Solution:-
- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.
E = Vo / D
E = 600 / 0.3
E = 2,000 V / m
- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:
Fe = E*q
Fe = (2,000 V / m) * ( 3*10^-5 C)
Fe = 0.06 N
- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:
Fnet = m*a
Where, a : The acceleration of the object/particle.
- The only unbalanced force acting on the particle is (Fe):
Fe = m*a
a = Fe / m
a = 0.06 / 0.0004
a = 150 m/s^2
- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:
s = C - A
s = ( 0.25 , 12 ) - ( 0.05 , 12 )
s = 0.2 m
- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:
vf^2 = vi^2 + 2*a*s
vf^2 = 0 + 2*0.2*150
vf = √60
vf = 7.746 m/s
- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:
vf - vi = a*t
t = ( 7.746 - 0 ) / 150
t = 0.0516 s
- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:
[tex]Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)][/tex]
- Where, The constant c : The capacitance of the capacitor.
- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:
[tex]E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\[/tex]
- Again, apply the Newton's second law of motion and determine the acceleration (a):
Fe = m*a
a = Fe / m
[tex]a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}[/tex]
- Where the acceleration is rate of change of velocity "dv/dt":
[tex]\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\[/tex]
- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:
[tex]c = \frac{A*eo}{d}[/tex]
Where, eo = 8.854 * 10^-12 .... permittivity of free space.
[tex]K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\[/tex]
- The differential equation turns out ot be:
[tex]\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\[/tex]
- Separate the variables the integrate over the interval :
t : ( 0 , t )
v : ( 0 , vf )
Therefore,
[tex]\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )[/tex]
- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:
[tex]vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s[/tex]
- The final velocity of particle while charging the capacitor would be:
vf = 7.264 m/s ... slightly less for the fully charged capacitor
A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is reduced, what happens to the width of the central bright fringe? Assume the angles involved remain small. The central bright fringe becomes narrower. The effect cannot be determined unless the distance between the slit and the screen is known. The central bright fringe becomes wider. The central bright fringe remains the same size.
Answer:
Explanation:
General guidance
Concepts and reason
The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.
Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.
Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.
Fundamentals
The condition for diffraction maximum in the case of single slit diffraction is as follows:
sin Θ=λ/α
Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .
The incorrect options are as follows:
• The central bright fringe remains in same size.
The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.
• The effect cannot be determined unless the distance between the slit and screen is known.
Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.
• The central bright fringe becomes narrower.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.
If the width of the slit increases, then the central bright fringe width also increases.
Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.
The condition for diffraction maximum is as follows:
sin Θ=λ/α
The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
Therefore,
• The central bright fringe becomes wider is correct.
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.
Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.
The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.
If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.
Answer
The applicable option when the width of the slit reduces is the central bright fringe becoming wider.
A coil of 43 turns in the shape of a rectangle with width 5 cm and length 10 cm is dropped from a position where magnetic field is 0 T to a position where the field is 0.5 T and is directed perpendicular to the plane of the coil. The displacement occurs in a time of 0.431 s. Calculate the resulting average emf induced in the coil
Answer:
Explanation:
Given that,
Number of turns
N = 43 turns
Dimensions of rectangle
Length L = 10cm = 0.1m
Width W = 5cm = 0.05
Then, area of rectangle
A = L×W = 0.1 × 0.05 = 0.005m²
Initial magnetic field
Bi = 0T
Final magnetic field
Bf = 0.5T
The displacement time
∆t = 0.431s
Induce emf ε?
EMF is given as
ε = —NdΦ/dt
Where magnetic flux
Φ = BA
Note A is constant
Then,
ε = —NdΦ/dt = —Nd(BA)/dt
ε = —N•A•dB/dt
ε = —N•A•(∆B/∆t)
ε = —N•A•(Bf - Bi) / (∆t)
ε = —43 × 0.005 × (0.5 - 0) /0.431
ε = —43 × 0.005 × 0.5 /0.431
ε = —0.249 V
The average induced EMF is 0.249V
Write the ordered pair that represents vec yz .Then find the magnitude of vec yz Y(- 4, 12), (1, 19)
Answer with Explanation:
We are given that
Y=(-4,12)
Z=(1,19)
We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.
Vector YZ=Z-Y=<1,19>-<-4,12>
Vector YZ=<5,7>
Magnitude of vector r
[tex]r=xi+yj[/tex]
[tex]\mid r\mid=\sqrt{x^2+y^2}[/tex]
Using the formula
[tex]\vec{YZ}=\sqrt{5^2+7^2}[/tex]
[tex]\vec{YZ}=\sqrt{25+49}=\sqrt{74} units[/tex]
Question 1
The orange line is measuring the
Answer:
Approximately 4.5 square miles
Explanation:
(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)
Final answer:
The 'orange line' referred to in the question likely represents data in a graph related to either the wavelength or frequency of orange light in the visible spectrum or a trend in statistical analysis. The exact answer depends on the specific context of the graph or data set from which the 'orange line' is derived.
Explanation:
The question 'The orange line is measuring the' seems to be related to analyzing data, most likely from a graph or chart that measures certain variables, potentially in the context of physics or mathematics dealing with light spectra or statistical analysis. Given the details provided, including mention of the visible light spectrum and specific wavelengths denoted by numbers, it appears that the 'orange line' may refer to a line on a graph that represents the wavelength or frequency of the color orange within the visible light spectrum.
However, the context provided is ambiguous. For instance, the chapter headings and the Conclusion provided suggests statistical analysis, perhaps in the field of queueing theory and how waiting times are affected by different systems, which implies a more mathematical or business-oriented question. In the field of mathematics, 'orange line' could also be a reference to a trend line or data line in a statistical plot.
4
An 80 kg skier stands at the top of a 40 meter slope. What is her gravitational potential energy
before she skies down the slope?
O 3200
O 31,360
19.63
O 15,680
Answer:
31,360J
Explanation:
Gravitation potential energy (gpe) is calculated from the formula mgh.
That implies, gpe = mgh
Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J
Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.
The gravitational potential energy of the skier at the top of the slope, calculated using the formula PE = mgh, is 31,360 joules.
Explanation:To find the gravitational potential energy of the skier, we use the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration because of gravity, and h is the height. The acceleration because gravity is 9.8 m/s².
So, we substitute these values into the formula:
PE = (80 kg) × (9.8 m/s²) × (40 m) = 31,360 joules.
Therefore, the gravitational potential energy of the skier before she skips down the slope is 31,360 joules.
Learn more about Gravitational Potential Energy here:https://brainly.com/question/23134321
#SPJ11
Global Precipitation Measurement (GPM) is a tool scientist use to forecast weather. Which statements describe GPM
Answer:
Correct answers:
Explanation:
• It is a satellite that collects data about rain and snow.
GPM is part of NASA's Earth Systematic Missions program and its function is to track precipitation from space with satellites and to provide accurate information of the time, place an amount of rain or snow anywhere in the world.
• Its orbit covers 90 percent of Earth's surface.
GPM has unique perspective on measuring precipitation from space because of their collected satellite retrievals.
• The sensors measure microwaves.
GPM Microwave Imager (GMI) is a sensor that observes the microwave energy emitted by the Earth and atmosphere, so it improves the accuracy of rain and snowfall estimates.
A T-shirt cannon is made of pipe within which a light spring, with k=100 N/m, can be compressed to launch the bundled shirt of mass 0.3 kg. Imagine the spring compressed 0.15 m and then released. Ignoring friction, calculate the launch velocity (speed at which the T-shirt leaves the spring) in two cases: A horizontal launch A vertical launch HTML EditorKeyboard Shortcuts
Answer:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Explanation:
The launch speed is calculated by means of the Principle of Energy Conservation:
[tex]U_{k} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = x \cdot \sqrt{\frac{k}{m} }[/tex]
[tex]v = (0.15\,m)\cdot \sqrt{\frac{100\,\frac{N}{m} }{0.3\,kg} }[/tex]
[tex]v \approx 2.739\,\frac{m}{s}[/tex]
The velocities for each scenario are presented herein:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Final answer:
The launch velocity of the T-shirt, when the spring with a spring constant of 100 N/m is compressed by 0.15 m, is calculated using conservation of energy. It turns out to be approximately 2.74 m/s for both the horizontal and vertical launches.
Explanation:
To calculate the launch velocity of the T-shirt from the cannon in both horizontal and vertical scenarios, we'll use the conservation of energy principle. This principle states that the potential energy stored in the compressed spring is converted into kinetic energy of the moving T-shirt.
Calculation of Launch Velocity (Horizontal and Vertical)
The energy stored in the compressed spring can be given by the formula:
[tex]E = 1/2 k x^2[/tex]
where E is the energy in joules, k is the spring constant, and x is the compression distance.
For the given spring with k = 100 N/m compressed by 0.15 m, the energy stored is:
[tex]E = 1/2 (100) (0.15)^2 = 1.125 joules[/tex]
Now we set this equal to the kinetic energy of the T-shirt upon release:
[tex]KE = 1/2 m v^2[/tex]
where KE is the kinetic energy, m is the mass of the T-shirt, and v is the velocity.
To find the launch velocity, we solve for v given the mass m = 0.3 kg and energy E = KE:
[tex]v = sqrt((2 * E) / m) = sqrt((2 * 1.125) / 0.3) = sqrt(7.5) = approximately 2.74 m/s[/tex]
This velocity will be the same for both the horizontal and the vertical launch as friction is ignored and the energy is purely transferred into the kinetic energy of the T-shirt.