Answer:
b) twice the energy of each photon of the red light.
Explanation:
[tex]\lambda[/tex] = Wavelength
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
Energy of a photon is given by
[tex]E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}[/tex]
Let [tex]\lambda_1[/tex] = 700 nm
[tex]\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}[/tex]
For red light
[tex]E_1=\dfrac{hc}{\lambda_1}[/tex]
For UV light
[tex]E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}[/tex]
Dividing the equations
[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1[/tex]
Hence, the answer is b) twice the energy of each photon of the red light.
Answer:
b) twice the energy of each photon of the red light.
Explanation:
Given:
wavelength of red light, [tex]\lambda_r=7\times 10^{-7}\ m[/tex]
wavelength of ultraviolet light, [tex]\lambda_{uv}=3.5\times 10^{-7}\ m[/tex]
We know that the energy of a photon is given as:
[tex]E=h.\nu[/tex] ..............(1)
where:
h = plank's constant [tex]=6.626\times 10^{-34}\ J.s[/tex]
[tex]\nu=[/tex] frequency of the wave
we have the relation:
[tex]\nu=\frac{c}{\lambda}[/tex] ...................(2)
From (1) and (2) we have:
[tex]E=\frac{h.c}{\lambda}[/tex]
Energy of photons for ultraviolet light:
[tex]E_{uv}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.5\times 10^{-7}}[/tex]
[tex]E_{uv}=5.6794\times 10^{-19}\ J[/tex]
Since the energy of photons is inversely proportional to the wavelength of the light hence the photon-energy of ultraviolet light is double of the photon-energy of the red light as the wavelength of the red light is twice to that of ultraviolet light.
A box slides down a slope described by the equation y=(0.05 x2)m, where x is in meters. If the box has x components of velocity and acceleration vX= -4m/s and aX= -1.2m/s@ at x = 3m, determine the corresponding y components of velocity and acceleration at this instant
Answer:
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
Explanation:
We are given that
[tex]y=(0.05x^2)[/tex]
Where x (in m)
x component of velocity, [tex]v_x=-4m/s[/tex]
x- component of acceleration,[tex]a_x=-1.2 m/s^2[/tex]
x=3 m
We have to find the y-component of velocity and acceleration at this instant.
Differentiate w.r.t. time
[tex]\frac{dy}{dt}=0.05\times 2x\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}=0.1x\frac{dx}{dt}=0.1xv_x[/tex]
By using the formula
[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
[tex]v_x=\frac{dx}{dt}[/tex]
Substitute the values
[tex]v_y=\frac{dy}{dt}=0.1(3)\times (-4)=-1.2m/s[/tex]
Again differentiate w.r.t. t
[tex]a_y=\frac{dv_y}{dt}=0.1v_x+0.1x\frac{d(v_x)}{dt}[/tex]
Substitute the values
[tex]a_y=0.1(-4)+0.1(3)(-1.2)=-0.4-0.36=-0.76m/s^2[/tex]
Where [tex]a_x=\frac{dv_x}{dt}[/tex]
Hence, the y- component of velocity and acceleration
[tex]v_y=-1.2m/s,a_y=-0.76m/s^2[/tex]
To find the y components of velocity and acceleration, differentiate the equation y = 0.05x^2 with respect to time. Plug in the given values to find vy and ay at x = 3m.
Explanation:To find the corresponding y components of velocity and acceleration at x = 3m, we need to differentiate the equation y = 0.05x^2 with respect to time. Differentiating y with respect to time gives vy = dy/dt = d/dt(0.05x^2) = 0.1x(dx/dt). Plugging in the given values dx/dt = -4m/s and x = 3m, we can find vy. Similarly, to find the y-component of acceleration, we differentiate vy with respect to time, which gives ay = dvy/dt = d/dt(0.1x(dx/dt)) = 0.1(dx/dt)(dx/dt) + 0.1x(d^2x/dt^2). Plugging in the given values dx/dt = -4m/s and d^2x/dt^2 = -1.2m/s^2, we can find ay.
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Suppose you want to move a big rock in your yard. It is about waist high and 4 feet long, has a volume of 1.2 m3 and a density of 2.5 g/cm3. How much does it weigh, in units of lbs?
Answer:
6613.87 lbs
Explanation:
1.2 m³ = 1200000 cm³
Mass = Density * Volume
M = (2.5 g/cm³) * 1200000 cm³ = 3000000 g
1 lb = 453.592 g
3000000 g * (1 lbs / 453.592 g) = 6613.87 lbs
The mass of the rock is 6613.87 lbs.
How do you calculate the Weigh?Mass is the quantity of matter in a physical body and the weigh is used to measure ascertain the heaviness of by or as if by a balance.
Given that volume of the rock is 1.2 meter cube and density is 2.5 gram/cm cube. The height of the rock is 4 feet. The mass can be calculated as given below.
Mass m = Density w [tex]\times[/tex] Volume v
Substituting the values in the above equation.
[tex]m = 2.5 \;\rm g/cm^3 \times 1.2\;\rm m^3[/tex]
[tex]m = 2.5 \times 1.2 \times 1000000[/tex]
[tex]m = 3000000\;\rm g[/tex]
The mass of the rock is 3000000 g. We know that
[tex]1\;\rm lb = 453.592\;\rm g[/tex]
[tex]1\;\rm g = \dfrac {1}{453.592} \;\rm lb[/tex]
Now convert the mass (grams) into mass (lbs).
[tex]m = \dfrac {1}{453.592}\times 3000000\;\rm lbs[/tex]
[tex]m = 6613.87 \;\rm lbs[/tex]
Hence we can conclude that the mass of the rock is 6613.87 lbs.
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How many meters will sound travel in 60 seconds if the temperature is 20℃ ? How far in an hour ? How far in 0.002 seconds ?
Answer:
a. 20652 m
b. 1239120 m
c. 0.688 m
Explanation:
First we have to find the speed of sound in air of temperature 20°C.
The speed of sound in a medium is dependent on the temperature of that medium. At any given temperature, the speed of sound is given as:
[tex]v(T) = v(T_{0}) + 0.61T[/tex]
where
v(T) = Speed of sound at any temperature T;
[tex]v(T_{0})[/tex] = Speed of sound at 0°C;
T = temperature of the medium.
The speed of sound in air of 0°C is 332 m/s, hence, in air of temperature 20°C, speed will be:
[tex]v(20) = 332 + (0.61 * 20)\\\\v(20) = 332 + 12.2\\\\v(20) = 344.2 m/s[/tex]
a. Distance traveled in 60 seconds:
distance = speed * time
distance = 344.2 * 60
distance = 20652 m
b. Distance traveled in 1 hour:
1 hour in seconds = 1 * 60 * 60 = 3600 seconds
distance = 344.2 * 3600
distance = 1239120 m
c. Distance traveled in 0.002 second:
distance = 344.2 * 0.002
distance = 0.688 m
A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.?
Final answer:
To find the displacement, subtract the distance traveled in one direction from the distance traveled in the opposite direction.
Explanation:
To solve this problem, we need to consider the concept of displacement. Displacement is a vector quantity that represents the change in position of an object. To find the displacement, we can subtract the distance traveled in one direction from the distance traveled in the opposite direction. In this case, the girl rides 5.4 km due east and then turns around and rides back to her starting point. So her displacement is 5.4 km - 5.4 km = 0 km.
The kinetic energy of a photoionized electron increases linearly with the wavelength of the ionizing radiation, but is independent of the intensity of the radiation. If False, explain in one sentence what makes the statement False.A. True
B. False
Answer: False
Explanation:
It is dependent on the intensity of the radiation i.e the photo electron should increase with the light amplitude.
A post is wrapped two full turns around with a belt. The tension in the belt is 7500 N by exerting a force of 150 N on its free end. Determine the coefficient of static friction between the belt and the post.
Answer:
Coefficient of friction is
Ū = 0.31
Explanation:
T2 = T1* e^(ūơ)
Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,
ū = coefficient of friction
Ơ = 2pai * N
Where N = number of turns = 2
Ơ = 4pai
7500 = 150e^(ū*4pai)
50 = e^(ū *4pai)
lin 50 = 4pai * ū
Ū = 3.91/4pai
Ū = 0.31
From whom should a departing VFR aircraft request radar traffic information during ground operations?
Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.
Explanation:
Final answer:
During ground operations, a departing VFR aircraft should request radar traffic information from ground control. Once airborne, the responsibility switches to tower control for traffic updates.
Explanation:
A departing VFR (Visual Flight Rules) aircraft should request radar traffic information from the ground control (also known as ground movement control) before taxiing. Ground control is responsible for the safety on the taxiways and runways, except the active runway. Once airborne or when holding short of the runway, the aircraft switches to tower frequency, at which point the aircraft may receive radar traffic updates relevant to their departure.
Technology utilizing a network of radio frequency waveguides is part of managing air traffic, which is crucial for maintaining a safe separation between aircraft. Furthermore, radar technology can provide directional information, and as noted in research such as Hasselmann, 1991, traffic information can be directly calculated from radar data even during the ground operations of VFR aircraft.
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one piece is 7.0 times the resistance of the other
Determine the length of the short piece.(% of length of the wire)?
Determine the resistance of the short piece?
Determine the resistance of the long piece?
Answer with Explanation:
Let r be the resistance of short piece of copper wire.
Resistance of copper wire=R=[tex]13\Omega[/tex]
Resistance is directly proportional to length.
If a wire has greater resistance then,the wire will be greater in length.
Therefore,resistance of long piece of wire=7r
Total resistance of copper wire=Sum of resistance of two piece of wires
[tex]r+7r=13[/tex]
[tex]8r=13[/tex]
[tex]r=\frac{13}{8}[/tex]ohm
Resistance of long piece of wire=[tex]7\times\frac{13}{8}=\frac{91}{8}\Omega[/tex]
Resistance of short piece of wire =[tex]\frac{13}{8}\Omega[/tex]
Resistivity of wire and cross section area of wire remains same .
Let L be the total length of wire and L' be the length of short piece of wire.
We know that
[tex]R=\frac{\rho L}{A}[/tex]=[tex]\frac{\rho}{A}L=KL[/tex]
[tex]\frac{R}{L}=K[/tex]
Where K=[tex]\frac{\rho}{A}[/tex]=Constant
Using the formula
[tex]\frac{13}{L}=\frac{\frac{13}{8}}{L'}[/tex]
[tex]\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}[/tex]
[tex]L'=\frac{L}{8}[/tex]
Length of short piece of wire=L'=[tex]\frac{L}{8}[/tex]
Length of long piece of wire=[tex]L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L[/tex]
% of length of short piece of wire=[tex]\frac{\frac{L}{8}}{L}\times 100=12.5%[/tex]%
The resistance of the short piece=[tex]\frac{13}{8}\Omega[/tex]
The resistance of the long piece=[tex]\frac{91}{8}\Omega[/tex]
The length of short piece wire is [tex]12.5[/tex] % of total wire length.
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex]
The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]
Resistance :Let us consider that resistance of short piece wire is r.
So that, resistance of long piece wire is [tex]7r[/tex].
Total resistance of wire is [tex]13[/tex] ohms.
[tex]7r+r=13\\\\8r=13\\\\r=\frac{13}{8}Ohm[/tex]
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex] The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]The resistance of wire is directly proportional to length of wire.Let us consider that total length of wire is 13L.
So that, length of short piece wire [tex]=\frac{13}{8} L[/tex]
length of short piece wire is,
[tex]=\frac{\frac{13}{8}L }{13L}*100 =\frac{1}{8} *100=12.5[/tex]
The length of short piece wire is 12.5 % of total wire length.
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The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension developed in the cable. The coefficient of kinetic friction between the crate and the plane is mk = 0.3.
Answer:
85 N
Explanation:
Given that crate mass = 20kg
Distance = 6m
Time = 3 seconds
Coefficient of kinetic friction = 0.3
We begin by calculating for acceleration
Which was gotten as 1.33 m/s sq
SEE THE ATTACHEMENT FOR DETAILS
To determine the tension developed in the cable, we consider forces acting on the crate, including the tension in the cable and the force of kinetic friction. We use Newton's second law of motion and equations for tension and force of kinetic friction to calculate the tension. By substituting values into the equations, we can solve for the tension.
Explanation:To determine the tension developed in the cable, we need to consider the forces acting on the crate. Since the crate is moving with constant acceleration, we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the forces acting on the crate are the tension in the cable and the force of kinetic friction. The tension in the cable can be calculated using the equation:
Tension = mass of the crate * acceleration + force of kinetic friction
Given that the mass of the crate is 20 kg and the acceleration is the change in velocity divided by the time taken (6 m / 3 s = 2 m/s), we can calculate the tension using the equation:
Tension = 20 kg * 2 m/s^2 + force of kinetic friction
We also need to determine the force of kinetic friction. The force of kinetic friction can be calculated using the equation:
Force of kinetic friction = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the crate, which can be calculated using the equation:
Normal force = mass of the crate * gravitational acceleration
Given that the coefficient of kinetic friction is 0.3 and the gravitational acceleration is 9.8 m/s^2, we can calculate the force of kinetic friction using the equation:
Force of kinetic friction = 0.3 * (20 kg * 9.8 m/s^2)
Now we can substitute this value back into the equation for tension:
Tension = 20 kg * 2 m/s^2 + (0.3 * (20 kg * 9.8 m/s^2))
Solving this equation will give us the tension developed in the cable.
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Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? ignore air resistance
A. When you hold the ball still in your hands after catching it
B. Just after the ball first leaves your hands.
C. At the instant the ball reaches its highest point.
D. At the instant the falling ball hits your hands.
E. When you hold the ball still in your hands before it is thrown.
Answer:
C. At the instant the ball reaches its highest point.
Explanation:
When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"
Remember, F = ma = m(v/t)
Since v = 0 at maximum height
F = m(0/t)
F = 0N
This shows that the force acting on the body is zero at the maximum height.
Kellie jogs 6.0 km in 54 minutes, then 1.0 km in 16 minutes. What is her average speed in km/min please show your work
Answer:
Average speed: 0.1 km/min
Explanation:
The average speed of a body is the ratio between the total distance covered and the time taken:
[tex]v=\frac{d}{t}[/tex]
where
d = distance covered
t = time taken
In this problem, we have:
- Distance covered is the sum of the two distances:
d = 6.0 + 1.0 = 7.0 km
- Time taken is the sume of the two times:
t = 54 + 16 = 70 min
Therefore, the average speed is
[tex]v=\frac{7.0}{70}=0.1 km/min[/tex]
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 244 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume that distilled water is an insulator.
(a) What is the charge on the plates before immersion?
What is the charge on the plates after immersion?
(b) What is the capacitance after immersion?
What is the potential difference after immersion?
(c) What is the change in energy of the capacitor?
?U = Uwater - Uair = 5 nJ
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
A 30kg rock is swung in a circular path and in a vertical plane on a .25m length string at the top of the path, the angular speed 12.0 rad/s. what is the tension in the string at that point?
A. 7.9 N.
B. 16 N.
C. 18 N.
D. 83 N.
Answer
given,
mass of the rock, m = 30 Kg
Length of the string, r = 0.25 m
angular speed of the rock, ω = 12 rad/s
Tension in the string at the top of the vertical circle = ?
At the top of the string the tension is
[tex]T_{top}=\dfrac{mv^2}{r} - m g[/tex]
we know v = r ω
[tex]T_{top}= mr\omega^2 - m g[/tex]
[tex]T_{top}= 30\times 0.25\times 12^2-30\times 9.8[/tex]
[tex] T_{top} = 786 N[/tex]
if the given mass is 0.3 Kg then tension in the rope
[tex]T_{top}= 0.30\times 0.25\times 12^2-0.30\times 9.8[/tex]
[tex]T_{top} = 7.86 N[/tex]
If the mass is 0.3 then the correct answer is option A.
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.3 km, but when the snow clears, he discovers that he actually traveled 8.4 km at 47o north of due east. (a) How far and (b) in what direction (south of due west) must he now travel to reach base camp
Answer:
[tex]s=5.79\ km[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Explanation:
Given:
distance of the person form the initial position, [tex]d'=8.4\ km[/tex]direction of the person from the initial position, [tex]47^{\circ}[/tex] north of eastdistance supposed to travel form the initial position, [tex]d=5.3\ km[/tex]direction supposed to travel from the initial position, due NorthNow refer the schematic for visualization of situation:
[tex]y=d'.\sin47^{\circ}-d[/tex]
[tex]y=8.4\times \sin47-5.3[/tex] ...............(1)
[tex]x=d'.\cos47^{\circ}[/tex]
[tex]x=8.4\times \cos47^{\circ}[/tex] .................(2)
Now the direction of the desired position with respect to south:
[tex]\tan\theta=\frac{y}{x}[/tex]
[tex]\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Now the distance from the current position to the desired position:
[tex]s=\sqrt{x^2+y^2}[/tex]
[tex]s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}[/tex]
[tex]s=5.79\ km[/tex]
Unlike a phonograph record that has a constant angular speed, a CD scans information at a constant linear speed (130 cm/s). Does a CD therefore rotate at a constant angular speed, or varying angular speed?
Answer: A CD rotate at a varying angular speed
A CD rotates at varying angular speed despite having a constant linear speed. This variation in rotation speed is necessary for the CD to operate efficiently.
Unlike a phonograph record, a CD rotates at varying angular speed even though it scans information at a constant linear speed due to the layout of data on the disc. The rotation speed decreases as the laser moves from the inside to the outside of the CD, maintaining a constant linear speed and adjusting the angular speed accordingly. This change in angular speed is essential for the CD to function effectively.
Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block 2. A force is applied to R2 and the blocks accelerate forward. Is the magnitude of the force exerted by the rope R1 on block 1 larger, smaller, or equal to the magnitude of the force exerted by R2 on block 2?
Final answer:
The force exerted by rope R1 on block 1 is equal to the tension in that rope, and if both blocks have the same acceleration and there are no other external forces, this tension is also equal to the force R1 exerts on block 2. However, the force applied through rope R2 on block 2 may be larger because it must account for the total mass and acceleration of both blocks.
Explanation:
When two blocks are connected by a rope and a force is applied to one of the blocks (via a second rope), the tensions in the ropes are subjects of interest in Physics, specifically classical mechanics. In this scenario, the force applied to rope R2 is used to accelerate both blocks 1 and 2. The tension in rope R1, which connects blocks 1 and 2, should be analyzed using a free-body diagram for each block.
Rope R1 exerts a force on block 1, and rope R2 exerts a force on block 2. The force exerted by rope R2 must be sufficient not only to accelerate block 2 but also to account for the force needed to accelerate block 1. Therefore, the force exerted by R2 on block 2 is equal to the sum of the forces necessary to accelerate both blocks. This total force must be greater than or equal to the force that R1 exerts on block 1, as rope R1 only has to provide the force needed to accelerate block 1. However, if we assume the acceleration of the two blocks is the same and the system is not experiencing any external forces other than the applied force, the tension in rope R1 will be equal to the force R1 exerts on block 1 because the massless ropes can only transmit the force equally along their lengths.
Therefore, it can be concluded that the magnitude of the force exerted by rope R1 on block 1 is equal to the magnitude of the tension on block 2 by rope R1, but the magnitude of the applied force through R2 on block 2 could be larger as it has to account for the combined mass of block 1 and block 2.
Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3.0 s to make one round trip. You construct another vertical oscillator with a mass 6 times as heavy and a spring 10 times as stiff. You take it to a planet where gplanet = 8.8 N/kg. You start it oscillating with amplitude 4.8 cm. How long does it take for the mass to make one round trip?
Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
Heat is a) the amount of thermal energy in an object. b) the energy that moves from a hotter object to a colder object c) a fluid-like substance that flows from a hotter object to a colder object. d) both A and B. e) both B and C.
Answer: Option (b) is the correct answer.
Explanation:
Heat is defined as the energy obtained by the molecules of an object that travels from a hotter body to a colder body.
For example, a metal spoon placed in a cup full of hot tea. Then transfer of heat will take place from the tea to the spoon.
And, this transfer of heat will continue till the temperature of both the objects become equal.
Thus, we can conclude that heat is the energy that moves from a hotter object to a colder object.
How many conditions does the NEC list whereby conductors shall be considered to be outside of a building or other structure?
Answer:
4 conditions
Explanation:
The National Electrical Code (NEC) provides safety guidelines for the installation of electrical wiring and electrical equipment in the United States. The purpose behind NEC is to standardize the safe electrical installation practices.
There is always a confusion among people to distinguish between inside or outside of a building or structure. Therefore, NEC has listed 4 conditions in the Article 230.6 where it has mentioned conductors to be considered outside a building or structure.
Conductors are considered outside a building when they are installed:
1. Under not less than 2 inches of concrete beneath a building or structure.
2. Within a building or structure in a raceway that is encased in no less than 2 inches thick of concrete or brick.
3. Installed in a vault that meets the construction requirements of Article 450, Part III.
4. In conduit under not less than 18 inches of earth beneath a building or structure.
The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u.
A) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving upstream and back downstream.
Express your answer in terms of the variables D, v, and u.
B) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving directly across the river and back.
Express your answer in terms of the variables D, v, and u.
Answer:
a) time needed to make a round trip = DV/v2 - u2
b) time needed to make a round trip = D/root ( v2 -u2)
Explanation:
The detailed explanation is as shown in the attachment
For a round trip moving upstream and downstream, the formula for the total time needed is D / (v - u) + D / (v + u). But, if the boat moves directly across the river and back, the total time needed is D / sqrt(v² - u²). This involves the concept of relative velocity.
Explanation:First, we need to understand the concept of relative velocity here. When the boat travels upstream, it moves against the current, so its net speed is (v - u). Downstream, it moves with the current at a net speed of (v + u).
A) The total time for a round trip moving upstream and downstream is the total distance divided by the speed in each direction. The time to travel a distance D upstream is D / (v - u), and the time to travel the same distance downstream is D / (v + u). So, the formula for the total time needed is D / (v - u) + D / (v + u).
B) If the boat is going directly across the river and back, it is moving perpendicular to the current. The boat makes the round trip at a constant speed of sqrt(v² - u²). Hence, the time needed for a round trip of total distance D is D / sqrt(v² - u²).
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A 58.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 5.00 m. What is the work done on the man by (a) the gravitational force and (b) the escalator
Answer
a) -2842 J
b) 2842 J
Explanation:
The escalator is moving the man of mass 58 kg till the height of 5 meters
So here;
m=58 kg
distance = height = h= 5 m
gravitational acceleration = g= 9.8 m/sec2
a) We know work is scalar product of force and displacement
i-e Work = F.S= FS cosθ, where θ is the angle between force vector and displacement vector
When escalator is moving up - the gravitational force is acting downward i- opposite to displcacement i-e angle between force and displacement is 180 degrees.
we have data
m= 58 kg
F=mg = 58×9.8=568.4 N
S=5 m
Work = F.S=FS cos 180°=568.4×5×(-1)= -2842 J
So the work done by gravitational force is -2842 J and -ve sign here indicates that distance is traveled in opposite direction of force.
b) When escalator is moving up the, force is exerted by the escalator equal to gravitational force and the displacement is in the same direction so the angle between force and displacement is 0 degrees
Work = F.S=FS cos (0)=568.4×5×1=2842 J
So work done by the gravitational force then will be 2842 J
So in both cases work is same but in opposite directions.
While surveying a cave, a spelunker follows a passage 190 m straight west, then 220 m in a direction 45.0 ∘ east of south, and then 270 m at 30.0 ∘ east of north. After a fourth unmeasured displacement, she finds herself back where she started. Use vector components to find the magnitude and direction of the fourth displacement.
Answer:
The magnitude of the displacement is 127.43m and the direction is 38°North of East
Explanation:
Final answer:
To find the magnitude and direction of the fourth displacement in a cave survey, vector components can be used. By breaking down the given displacements into their north and east components, calculating the net displacements, and using the Pythagorean theorem and trigonometry, the magnitude and direction of the fourth displacement can be determined.
Explanation:
To find the magnitude and direction of the fourth displacement, we can use vector components. First, let's break down the given displacements into their north and east components:
Displacement 1: 190 m west (0 north, -190 east)
Displacement 2: 220 m at 45.0° east of south (-220 sin(45°) north, 220 cos(45°) east)
Displacement 3: 270 m at 30.0° east of north (270 sin(30°) north, 270 cos(30°) east)
Next, we can add up the north and east components separately to find the net north and east displacements. Lastly, we can use these net displacements to calculate the magnitude and direction of the fourth displacement using the Pythagorean theorem and trigonometry.
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back and releases it when the seat is 1.09 m above the ground. The acceleration of gravity is 9.8 m/s 2 . How fast is Betty moving when the swing passes through its lowest position? Answer in units of m/s.
Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = [tex]\frac{1}{2} *m*vf^{2}[/tex] (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
[tex]vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s[/tex]
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.
Final answer:
Betty is moving at a speed of approximately 2.86 m/s when the swing passes through its lowest position.
Explanation:
To calculate the speed at the lowest position of Betty on the swing, we can use the principle of conservation of mechanical energy. At the highest point, all of Betty's potential energy is converted into kinetic energy. So we can write:
Initial potential energy = Final kinetic energy
mgh = 0.5mv^2
where m is the mass of Betty, g is the acceleration due to gravity, h is the initial height, and v is the velocity at the lowest point.
Plugging in the values, we get:
(400 N)(0.21 m) = 0.5(400 N)v^2
After simplifying and solving for v, the velocity at the lowest point is approximately 2.86 m/s.
The displacement of thoughts, feelings, fears, wishes, and conflicts from past relationships onto new relationships is called _____.
The displacement of thoughts, feelings, fears, wishes, and conflicts from past relationships onto new relationships is called transference.
A Ford Taurus driven 12,000 miles a year will use about 650 gal of gasoline compared to a Ford Explorer that would use 850 gal. About 19.8 lbm of CO2, which causes global warming, is released to the atmosphere when a gallon of gasoline is burned. Determine the extra amount of CO2 production a man is responsible for during a 5-year period if he trades in his Taurus for an Explorer.
Answer:
19 800 lbm of carbon dioxide more.
Explanation:
Taurus
Amount of gasoline used in 5 years = 650 ga/year * 5 years = 3250 ga
amount of carbon dioxide released = 19.8 lbm/ga * 3250 ga = 64 350 lbm
Explorer
Amount of gasoline used in 5 years = 850 ga/year * 5 years = 4250 ga
amount of carbon dioxide released = 19.8 lbm/ga * 4250 ga = 84 150 lbm
Extra amount of CO2 released = 84 150 lbm - 64 350 lbm = 19 800 lbm
The extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.
Taurus
Amount of gasoline used in 5 years =
[tex]\bold {650\ gallon/year \times 5 = 3250\ gallon}[/tex]
Amount of carbon dioxide released =
[tex]\bold {19.8\ lbm/gallon \times 3250 = 64 350 lbm}[/tex]
Explorer
Amount of gasoline used in 5 years =
[tex]\bold {850\ gallons/year \times 5 = 4250\ gallons }[/tex]
Amount of carbon dioxide released =[tex]\bold { 19.8\ lbm/gallons \times 4250 gallons = 84 150\ lbm}[/tex]
Extra amount of [tex]\bold {CO_2}[/tex] released =
[tex]\bold {= 84 150 - 64 350 = 19 800\ lbm}[/tex]
Therefore, the extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.
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As magnification increases, the area of the field of view _______, the depth of the field of view _________, the working distance ______, and the amount of light required ____________.
Answer:
As magnification increases, the area of the field of view increases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
Area of the field of vision: As magnification is increased, the field of view narrows, and you can see less of the object or scene that you are watching. The area you can see reduces as a result of the field of view is smaller.
Depth of field: The range of distances inside an object or scene that simultaneously seem in focus is referred to as depth of field. The depth of field tends to get shallower as magnification gets bigger. This makes it more difficult to simultaneously examine objects at various depths because only a limited range of distances will ever be in focus.
The working distance is the separation between the thing being observed and the objective lens (or magnifying device). The working distance often reduces as magnification rises. To keep the object in focus, you must move it closer to the lens or microscope.
Higher magnification frequently necessitates additional light in order to preserve image quality and visibility. The features become more noticeable when you zoom in on a smaller region, although they might need more light to be clearly visible. At greater magnifications, insufficient light can produce a hazy or dim image, therefore more light would be required.
Hence, As magnification increases, the area of the field of view decreases, the depth of the field of view decreases, the working distance decreases, and the amount of light required increases.
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A toy roller coaster cart has a mass of 2kg. The track of the roller coaster has a semicircular valley with a radius of 1.2 meters. If the cart has a speed of 8 m/s at the very bottom of the valley, find the amount of centripetal force necessary to complete the motion
Answer:106.6N
Explanation:
Centripetal force is a force that cause a body to move along a circular path in a field. The body possesses a mass and velocity(since it's moving) and also radius alone the circular path on which it moves.
Given mass (m) = 2kg, velocity (v) = 8m/s and radius (r) = 1.2m
Centripetal acceleration(a)= v²/r
Centripetal force = ma = mv²/r
Substituting this values in the formula above we have;
Centripetal force = 2×8²/1.2m
Centripetal acceleration = 128/1.2
= 106.6N
The greenhouse effect plays a major role in the climate of a planet. Visit the Greenhouse Effect AstroTour (Links to an external site.), and use what you learn there to choose all the statements below that are correct. Note that 273 K = 0°C = 32°F.
Answer:
If there were no greenhouse effect, liquid water would not exist on the surface of the Earth.
The Earth has reached thermal equilibrium, emitting the same amount of energy into space as it absorbs from the Sun.
The more carbon dioxide there is in an atmosphere, the stronger the greenhouse effect will be.
Explanation:
Without greenhouse gases like carbon dioxide, the entire Earth would be well below the freezing temperature of water. These gases slow the radiation of infrared light into space, warming the planet until it reaches thermal equilibrium.
The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It involves the trapping of heat by greenhouse gases in the atmosphere, which raises the Earth's surface temperature. Without the greenhouse effect, the Earth would be much colder.
The greenhouse effect is a natural phenomenon that plays a major role in the Earth's climate. It is similar to the heating of a greenhouse or a car left in the sun with the windows rolled up. The greenhouse gases in the atmosphere, such as carbon dioxide and methane, trap heat from the sun, preventing it from escaping back into space. Without the greenhouse effect, the Earth's average surface temperature would be well below freezing and Earth would be locked in a global ice age.
Greenhouse gases like carbon dioxide and water vapor absorb and emit radiation, which is an important factor in the greenhouse effect. Approximately half of the radiation from the sun passes through these gases in the atmosphere and strikes the Earth. This radiation is converted into thermal radiation on the Earth's surface, which is then partially reflected back to the Earth's surface by greenhouse gases. The more greenhouse gases there are in the atmosphere, the more thermal energy is reflected back to the Earth's surface, causing it to heat up.
The greenhouse effect is responsible for raising the Earth's surface temperature by about 23 °C. Changes in global climate due to the greenhouse effect can lead to more intense storms, changes in precipitation patterns, reduction in biodiversity, and rising sea levels.
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A student who weighs 550 N is wearing a backpack that weighs 80 N. The student is standing still on level ground. Give your answers to the following questions in newtons. Remember, enter only the number; don't try to enter the units.a) What is the magnitude of the net force on the student?
b) What is the magnitude of the contact force on the student by the backpack?
c) What is the magnitude of the contact force on the student by the ground?
Answer:
Explanation:
a) The magnitude of the net force on the student = 0 N since the student is standing still on level ground and upward reaction force = downward force.
b) the magnitude of the contact force on the student by the backpack = 80 N since the student was backing the backpack
c) the magnitude of the contact force o the student by the ground = 550 N + 80 N = 630 N reactional force on the student
A ball is dropped and bounces off the floor. Its speed is the same immediately before and immediately after the collision. Which of the following is true about the collision between the ball and the floor?
a) The ball's momentum is conserved.
b) The ball's momentum changes direction but not magnitude.
c) The ball's momentum changes magnitude but not direction.
d) The impulse delivered to the ball by the floor is zero.
e) The impulse delivered to the floor by the ball is zero.
Answer:
option B.
Explanation:
The correct answer is option B.
when the ball drops, the velocity of the ball before the collision is v
After the collision, the velocity of the ball is the same but in the opposite direction.
Impulse delivered to the ball and the floor, in this case, is not zero.
The magnitude of the momentum remains the same but the direction of the ball changes.
The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.
What is momentum?The term momentum is physics is the product of mass and velocity. Momentum is a vector quantity hence when discussing momentum, we have to take the magnitude and direction into account.
The true statement about the collision between the ball and the floor is that the ball's momentum is conserved.
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