Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
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A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
Jack weighs 170 lbs and is 72 inches tall. He is pulling horizontally on a door handle situated at his shoulder height. Actually, it is his body weight and lean that creates this pulling action (a hint). His center of mass while standing erect is 61 percent of his body height, measured from the floor upwards. The door handle is 60 inches above the ground, and again he is pulling purely horizontally on this handle.
If Jack's lean angle is 20 degrees and he is leaning back - pivoting about his heels, how much force does he apply to the door handle?
Include units in your answer, lbs.
Express your answer to the nearest 0.1 lbs.
Answer:
He is pulling horizontally on a door handle situated at his shoulder height. ... His Center Of Mass While Standing Erect Is 61 Percent Of His Body Height, Measured ... Actually, it is his body weight and lean that creates this pulling action (a hint).
72ibs
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
A car travels 150 kilometers west in 3 hours. What is its average velocity?
Your answer:
150 km/hr
50 km/hr
50 km/hr west
150 km/hr west
Answer:
C= 50km/hr west
Explanation:
150/3= 50
Because it asks for velocity, make sure to include the direction as well.
PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 120oC while the other side contains 1 m3 of He gas at 500 kPa and 40oC. Assume the piston is made of 8 kg of copper initially at the average temperature of the two gases on both sides. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move
Answer:
The answer is "[tex]\bold{83.8^{\circ} \ C}[/tex]".
Explanation:
Formula for calculating the mass in He:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg[/tex]
Formula for calculating the mass in [tex]N_2[/tex]:
[tex]\to m = \frac{PV}{RT}\\[/tex]
[tex]= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg[/tex]
by using the temperature balancing the equation:
[tex]T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}[/tex]
[tex]= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C[/tex]
something that orbiys other things in space
Answer: well we all orbit the sun all the planets do so the
SuN
Explanation: two words common sense
What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?
Answer:
speed = 10.1215 m/s
Explanation:
speed = distance / time
speed = 100 / 9.88 = 10.1215 m/s
Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C. (drag on jet / drag on transport)
Answer:
[tex]2.267[/tex]
Explanation:
Drag force is given by
[tex]F=\dfrac{1}{2}\rho Av^2C[/tex]
C = Drag coefficient is constant
A = Area is constant
[tex]v_1[/tex] = Velocity of the passenger jet = 1200 km/h = [tex]\dfrac{1200}{3.6}\ \text{m/s}[/tex]
[tex]v_2[/tex] = Velocity of the prop plane = [tex]\dfrac{1}{4}v_1[/tex]
[tex]\rho_1[/tex] = Density of the air where the jet was flying = [tex]0.38\ \text{kg/m}^3[/tex]
[tex]\rho_2[/tex] = Density of the air where the prop plane was flying = [tex]0.67\ \text{kg/m}^3[/tex]
[tex]F\propto \rho v^2[/tex]
[tex]\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267[/tex]
The ratio of the drag forces is [tex]2.267[/tex].
A 14.0-g wad of sticky clay is hurled horizontally at a 90-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact
Answer:the speed of the clay immediately before impact =72.58m/s
Explanation:
Given that
mass of the stick clay, M₁= 14.0 g = 0.014 kg
mass of the block ,M₂= 90 g = 0.09 kg
Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg
Also, distance, s = 7.50 m
coefficient of friction μ= 0.650
Acceleration due to gravity ,g = 9.8 m/s²
Using the Work- Energy theorem,
change in kinetic energy = work done
final kinetic energy(K₂) - initial kinetic energy(K₁) = force, F x coefficient of friction, μ x distance,s
The final kinetic energy is zero because after the impact, the block with the clay comes to a stop after 7.50m
kinetic energy =Work done
0.5 x m x v²=coefficient of friction, μ x force(F) x distance,s(Since force = m g )
0.5 x m x v²= μ x m x g x s
0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5
v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104
v²==95.55
V = 9.77 m/s
Using the conservation of momentum formulae where
M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V
Since V₂ which is the velocity of block is zero as the block is initially at rest, We now have that
M₁ V₁ = (M₁ + M₂ ) V
0.014 kg x V₁ = 0.104 x 9.77
V₁=0.104 x 9.77 / 0.014
V=72.58m/s
The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the
Equator bring cooler waters to some regions, while currents flowing away from the Equator bring warmer waters to other regions.
North
British
Isles
Askan
North Atlantic
Azor
U.S.A
California
Gulf Stream
Loop
n
Canbean
North Equatorial
North Equatorial CC
North Fuatorial
Equator
South Equatorial
Not
South Equatorial
Image courtesy of NOAA
Judging from the map, which region probably has cooler summers than it would without the effect of a nearby ocean current?
A the Central U.S.
B. the British Isles
C. the U.S. East Coast
D. the US West Coast
Answer:
d
Explanation:
the US West Coast region probably has cooler summers than it would without the effect of a nearby ocean current.
what is ocean current ?
ocean current can be defined as the horizontal movement of seawater which is produced by gravity, wind, and water density, it play an major role in the determination of climates of coastal regions.
The movement of ocean water is continuous which can be up three types such as Waves, Tides, Currents
The streams of water which flow continuously on the ocean surface in specific directions are called ocean currents, it affect the temperature of ocean water as Warm ocean currents increase the temperature whereas Cold ocean currents decrease the temperature.
The magnitude of the ocean currents is about few centimeters per second to as much as 4 metres per second and the intensity of the ocean currents generally decreases with increasing depth.
There are two types of ocean currents such as Warm Ocean Currents
and Cold Ocean Currents
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an object of 5kg is attached to a rope of length 4m is Rotating horizontally at 8m/s horizontally 20m above the ground if the rope is suddenly cut what is the horizontal distance travelled by the object? Please guys help
Answer:
16 meters
Explanation:
When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.
First we need to find how much time de object take to reach at the ground.
VERTICALLY EQUATION:[tex]h(t)=h-v*t-\frac{g}{2} t^{2} \\[/tex]g=acceleration of gravity=10m/s²
v= vertical velocity =0m/s
h=vertical altitude =20m
We will find t such that h(t)=0
[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]
HORIZONTALLY EQUATION:*horizontally we do not have acceleration[tex]D(t)=v*t[/tex]v=horizontal velocity
D(t=2) is the horizontal distancetravelled by the object:
[tex]D(2)=8*2\\\\D(2)=16m[/tex]
two small identical conducting spheres have charges of 2.0x10-9C and - 0.5x109 C respectively when they are placed 4cm apart, what is the force between them? If they are brought into contact and then separated by 4cm, what is the force between them?
Answer:
6
Explanation:
nothingnsbejejjdbsbzbawkje
Scientists have investigated how quickly hoverflies start beating their wings when dropped both in complete darkness and in a lighted environment. Starting from rest, the insects were dropped from the top of a 50 - cm tall box. In the light, those flies that began flying 200 m s after being dropped avoided hitting the bottom of the box 87 % of the time, while those in the dark avoided hitting only 25 % of the time.
Required:
a. How far would a fly have fallen in the 200 ms before it began to beat its wings?
b. How long would it take for a fly to hit the bottom if it never began to fly? In seconds.
Answer:
Explanation:
a )
Hoverfly will fall with acceleration equal to g .
Initial velocity of fall of hoverflies u = 0
displacement ( vertical ) h = ?
time t = 0.2 s
acceleration due to gravity g = 9.8 m / s²
h = ut + 1/2 g t²
= 0 + .5 x 9.8 x .2²
= .196 m
= 19.6 cm
b )
Time taken to fall by 50 cm or 0.5 m under free fall from initial position .
.5 = 0 + .5 x 9.8 t²
t² = .1020
t = .319 s = 319 ms .
I need help on a homework question.
Chris shines a white light onto a surface, and the surface appears to be green. What color will the surface appear if he shines a combination of blue and green light on the surface?
A. brown
B. white
C. green
D. black
Answer:
green
Explanation:
the the board is going to be green
Answer:
The surface will appear green if he shines the combination of blue and green light.
Explanation:
When Chris shines white light on the surface it appears green it means the surface is scattering only green light.it is called scattering of light.The material of the surface is absorbing rest of the colors.What is scattering of light ?When light passes through a medium , the particles of the medium or surface allows to reflect only a particular color ( wavelength), this phenomenon is called scattering of light.
Hence the surface will appear green .
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2. The storage in a river reach at a specified time is 3 hectare-meters. At the same instant, the inflow to the reach is 15 m3/s and the outflow is 20 m3/s. One hour later, the inflow is 20 m3/s and the outflow is 20.5 m3/s. Determine the change in storage in the reach that occurred, during the hour. What is the storage at the end of the hour? (1 hectare = 10000 m2).
Answer:
The change in storage in the reach that occurred, during the hour = 43200 m3
The storage at the end of the hour = 129600 m3
Explanation:
Given
Storage = 3 hectare-meters
Inflow volume = 15 m3/s
Outflow volume = 20 m3/s
Storage in one hour of change
Inflow volume = 20 m3/s
Outflow volume = 20.5 m3/s
Outflow volume - Inflow volume = 0.5 m3/s = 0.5 * 24*60 * 60 = 43200 m3
Storage at the end of one hour
43200 m3 + (20-15) * 24*60 * 60 = 129600 m3
In picture 1, heat is flowing from the ____ to the _____ In picture 2, heat is flowing from the _______ to the ____
Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid
Explanation:
I don't know if I answered correctly, if not I can provide another answer
Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus a distance of 1.30 meters, how much work does the tension force do on Magnus
Answer:
Workdone = -2730 J
Explanation:
Formula for workdone is;
W = Force × Displacement
Now, according to Newton's 3rd law of motion, to every action, there is an equal and opposite reaction.
In the question given, we are told that a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus. This means that the force exerted by the rope on Magnus acts in an opposite direction to that which Magnus does to the rope.
Therefore, the force will be in the negative direction.
So;
Workdone = -2100 N × 1.3 m
Workdone = -2730 J
The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight
Answer:
D. Weight
Explanation:
Hope that helps:)
help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposing each other) in a circuit. If you wanted to add in a capacitor to charge it from the batteries, would you be able to get more charge onto the capacitor or less charge, than if there was only one battery. (hint: start this problem by aligning the batteries positive to negative, and think of it from conservation of energy perspective).
Answer:
Answer is explained in the explanation section below.
Explanation:
This question is very basic and easy. The answer to this question is:
Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.
Reasoning:
If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.
As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.
What are some technological limitations that currently prevent humans from traveling to distant planets?
Answer:
Propulsion system, antigravitational tech
Explanation:
Fuel is extremely inefficient and expensive not to mention it weighs a lot. You really only need to reach escape velocity to leave earth. The rest is just a little amount of boosting to alter course and slow down for landing. I couldn't really think of much. Once we have an antigravitational system then you could say the whole rocket is holding you back because the design would be different. Nobody really knows how to defy gravity but that would be a technolgical limitation for sure.
how does the strength of the forces that hold the basic particles of a substance together relate to the temperature at which the substance changes state
The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is?
Answer:
h over 2 dg
Explanation:
brainliest!!!!!!!
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g
Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
Do you ever have a sensation of loneliness?
Answer:
nope
Explanation:
sige bigyan kitang Happy pills
PLEASE HELP! WILL GIVE BRANILIEST TO FIRST REAL ANSWER If one marble is rolling three times as fast as a second marble of the same mass, the kinetic energy of the first marble is how many times larger when compared to the kinetic energy of the second marble?
a) 4
b) 9
c) 6
d) 3
(i already know its not 3)
Answer:
9
Explanation:
Select the correct answer.
The oceanic Nazca plate is being subducted beneath the continental South American plate. Which type of plate boundary is this?
OA continental-oceanic convergent
ОВ. oceanic-oceanic convergent
OC divergent
OD. strike-slip
ОЕ.
transform
Reset
Next
Answer:
A. continental-oceanic convergent
Explanation:
I knew it couldn't be B because it's oceanic and continental, not oceanic and oceanic.
Next, I noticed the word convergent, which implies "coming together" to me.
I looked it up and noticed the term convergent referred to a plate boundary where a plate slips under (subducted) another, so I knew it was A.
Hopefully, this helps you understand the question better. Have a great day!
Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not equally sensitive to all frequencies of sound. In other words, a particular noise may sound louder or softer depending on the frequency of the sound wave being transmitted. Because of this variation, scientists have defined a unit of loudness, called a phon, to represent the intensity of sound waves with a frequency of 1000 Hz. A 60-phon sound is one that is perceived by the human ear to have the same loudness as a sound wave with an intensity of 60 dB and a frequency of 1000 Hz.
Required:
a. At approximately what frequency do most people perceive the least intense sounds? Answer numerically in hertz to two significant figures.
b. Normal conversation has a sound level of about 60 dB. How many times more intense must a 100-Hz sound be compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness?
Answer:
20 Hz
15.8 times
Explanation:
A
Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.
B
A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness
It's time to get a little more specific. Based on the velocity (Vx) graph for the car and the velocity data in the table, divide the total
motion of the car into rough time periods that tell a different "chapter" of the story for this car trip. In each of these time
periods, the car's velocity will be notably different from the previous period. Enter a brief description of the car's motion in each
period. The first one is done for you. Use it as an example to identify and describe the remaining time periods. Note: You can
define as many periods as you think appropriate.
s
B
1
U X
X х.
Font Sizes
А • А
E
E 를 들
E 3
Numbered list
Time Period
Motion Description
0.2 - 4.6 seconds increasing speed in positive direction
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
