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].
The name of the SI unit for magnetic field strength, such as that created around a current-carrying wire, is the
.
The name of the SI unit for magnetic field strength is Tesla. Magnetic fields are formed by moving electric charges.
What is Magnetic field strength?Magnetic field strength alludes to an actual amount that is utilized as one of the essential proportions of the power of the attractive field.
The SI unit of attractive field is tesla (T). 1 Tesla is defined as the magnetic field that carries 1C charge at the speed of 1m/s which is perpendicular to the force of 1 N.
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A child holds a sled at rest on frictionless snow covered hill. if the sled weighs 77N,find the force T exerted by the rope on the sled and the force n exerted by the hill on the sled
Answer:62
Explanation:
The weight of the sled is 77 N. The force by the hill on the sled is equal to its weight that is 77 N. Then the tension force exerted by the rope on the sled is being 77N sin θ, where θ be the angle of inclination.
What is force?Force is an external agent acting on an object to change its motion or to deform it. There are various kinds of force like magnetic force, tension force, frictional force, gravitational force etc.
The weight that an object experience on earth is due to the gravitational force. The force that is exerted by a rope on an object is tension force since it is pulling from a side.
The normal force by the hill on the sled is equal to its weight that is 77 N. The tension force on the sled by the rope is dependent on the angle of inclination θ. If know the angle we can find T by the equation:
T = 77 sin θ.
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To fully describe velocity you must have a _____
A. Magnitude and unit
B. Speed and unit
C. Average speed and position
D. Magnitude and direction
is a step in the scientific method. The step that follows this step involves forming
Answer: read this hope this helped
Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.
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.
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.
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 .
Read it also for better understanding:-
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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
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.
Write the laws of reflection
Answer:
i) Angle of Reflection = Angle of Incident
ii) Total Angle = 2x Angle of reflection/incidence
iii) The incident ray, reflected ray and the point of incidence lie in the same plane
Hope it helped u,
pls mark as the brainliest - Ayaan707 for help :) -
^_^
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 .
what is the mass of an object that is accelerating at a rate of 25m/s^2 and using 15 N of force
Answer:
mass = 0.6 kg
Explanation:
Given
Acceleration (a) = 25 m/s²
Force (F) = 15 N
Mass (m) = ?
We know
F = M * a
15 = m * 25
m = 15 / 25
m = 0.6 Kg
Hope it will help :)
The mass of an object that is accelerating at a rate of 25 m/s^2 and using 15 N of force is 0.6 Kg.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question,
Acceleration (a) = 25 m/s²
Force (F) = 15 N
Mass (m) = ?
We know
F = M * a
15 = m * 25
m = 15 / 25
m = 0.6 Kg
The mass of an object that is accelerating at a rate of 25 m/s^2 and using 15 N of force is 0.6 Kg.
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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.
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
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]
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]
Galileo
o did not believe friction existed
o believed that friction stopped objects in motion
o believed that friction kept objects in motion
О
assumed that in a frictionless environment objects would never move
Answer:
object would move but it could be difficult to slow down or stop.
Do you ever have a sensation of loneliness?
Answer:
nope
Explanation:
sige bigyan kitang Happy pills
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
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
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
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
ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law say about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.
Answer:
A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.
(Hope this helps! Btw, I am the first to answer.)
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
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!
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 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
