Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, [tex]m=1.01\times 10^7\ kg[/tex]
Force applied on the tracks, [tex]F=7.5\times 10^5\ N[/tex]
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s[/tex]
So, the required time is 300.3 seconds.
If you were to stand in the exact center of a rotating disc, you would only have what kind of
speed?
Tangential speed
Increasing speed
Linear speed
Rotational speed
Answer:
Tangential speed or Rotational speed
Integrate your expressions for dEx and dEy from θ=0 to θ=π. The results will be the x-component and y-component of the electric field at P
.
Express your answers separated by a comma in terms of some, all, or none of the variables Q
and a and the constants k and π.
Answer:
hello your question is incomplete below is the missing part
Ex = 0
Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]
Explanation:
Attached below is a detailed solution showing the integration of the expression dEx and dEy from ∅ = 0 to ∅ =π
Ex = 0
Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]
An equipotential surface that surrounds a point charge q has a potential of 536 V and an area of 1.20 m2. Determine q.
Answer:
q = 1.84×10^-8coulombs
Explanation:
Surface area = 4πr²
r is the distance
1.2 = 4(3.14)r²
1.2 = 12.56r²
r² = 1.2/12.56
r² = 0.0956
r = √0.0956
r = 0.309m
Get the charge C
V = kq/r
536 = 9.0×10^9q/0.309
536×0.309 = 9×10^9q
165.73 = 9×10^9q.
q = 165.73/9×10^9
q = 1.84×10^-8coulombs
Which statement is correct?
A. If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
B. When the electric field is zero at a point, the potential must also be zero there.
C. If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
D. If the electric potential at a point in space is zero, then the electric field at that point must also be zero.
Answer:
The answer is "Choice C ".
Explanation:
The relationship between the E and V can be defined as follows:
[tex]\to E= -\Delta V[/tex]
Let,
[tex]\to E= \frac{\delta V}{\delta x}[/tex]
When E=0
[tex]\to \frac{\delta V}{\delta x}=0[/tex]
v is a constant value
Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.
Static Friction
Now let’s examine the static case. Remain on the “Force graphs” tab at the top of the window. Make sure the box labeled “Ffriction” is checked at the left of the screen, this will allow us to measure to force of friction experienced by an object as it slides down the ramp.
Draw a free body diagram for an object sitting on the incline at rest, assuming the incline is at the maximum angle BEFORE the object starts to move. Be sure to include friction and stipulate whether it is kinetic or static.
How much time does it take a dropped object to fall 180 m on Earth?
18 s
36 S
10 s
6s
Answer:
6s
Explanation:
Assume it is dropped from rest and the gravitational acceleration is 10
By the equation of motion under constant acceleration:
[tex]s=ut+\frac{1}{2} at^2[/tex]
180 = (0)t+10(t^2)/2
t = 6 or -6 (rejected)
t = 6 s
- .
?
y
(っ◔◡◔)っ ♥ chose the answer with the question marks ♥
Answer:
okay I'm a bit confused but I like the little emoji dudw
Answer:
?
Explanation:
.
A boy of mass 60 kg is sledding down a 70 m slope starting from rest. The slope is angled at 15° below the horizontal. After going 20 m along the slope he passes his friend of mass 50 kg, who jumps on the sled. They now move together to the bottom of the slope. The coefficient of kinetic friction between the sled and the snow is 0.12. Ignoring the mass of the sled, find their speed at the bottom.
A particle moves along the x-axis according to the equation (x=14-7t+t^2+t^3 ), where (x) in meter and (t) in seconds. At (t=7 sec) Find (a) The position of the particle (b) It’s velocity (c) It’s acceleration
Answer:
jjnn ok jjjmkkmmkijnnkko
2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
The screw does 65 J of work. What is the efficiency of the screw? Show your
work. Hellpppp
Answer:
42,250
Explanation:
It goes inside=
Displacemt
It does work=
Work done
To find efficiency of jule we do=
Dicplacement × Work done
650 × 65
42,250
Please mark me as a brainlist
Mr. Voytko wants to know how high in meters he can lift an 0.3 kg apple with 7.35 joules?
Answer:
the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
Explanation:
Given;
energy of Mr. Voytko, E = 7.35 J
mass of the apple, m = 0.3 kg
Apply the principle of conservation of energy.
Energy of Mr. Voytko = Potential energy of the apple due to its height above the ground.
E = mgh
where;
h is the height above the ground through Mr. Voytko lifted the apple.
g is acceleration due to gravity = 9.8 m/s²
h = E / (mg)
h = 7.35 / (0.3 x 9.8)
h = 2.5 m
Therefore, the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
A ball is thrown straight up into the air. Which of the following best describes the energy present at various stages?
There is more energy at the top of the ball's path than there is at the bottom.
The total amount of energy varies, with more energy at the bottom and less at the top of the path.
At the very top, most of the energy is potential and just before it hits the ground, most of the energy is kinetic.
At the very top, most of the energy is kinetic and just before it hits the ground, most of the energy is potential.
Answer:
Uhh 2 one
Explanation
.................,,,,,,,,,,,
Answer:
B
Explanation:
Motion is movement, the teacher's movement is motion
What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 cm from the other) is 293 V?
Answer:
E = 4156.02 Vm⁻¹
Explanation:
The magnitude of the uniform electric field between the plates can be given by the following formula:
[tex]E = \frac{\Delta V}{d}\\[/tex]
where,
E = Electric field strength = ?
ΔV = Potetial Difference = 293 V
d = distance between plates = 7.05 cm = 0.0705 m
Therefore,
[tex]E = \frac{293\ V}{0.0705\ m}\\\\[/tex]
E = 4156.02 Vm⁻¹
A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some solar panels on the roof of a building. Which form of energy to collected by the solar panels?
A. Wind
B. sound
C. Magnetic
D. Light
Calculate the magnitude of the gravitational force exerted by Mercury on a 70 kg human standing on the surface of Mercury. (The mass of Mercury is 3.31023 kg and its radius is 2.4106 m.)
Answer:
2.66×10⁻⁹ N.
Explanation:
From the question,
Applying newton's law of universal gravitation,
Fg = GMm/r²............................... Equation 1
Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury
Given: M = 3.31023 kg, M = 70 kg, r = 2.4106
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute these values into equation 1
Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)
Fg = 2.66×10⁻⁹ N.
A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow direction) of 17.5 m and a chord (the length parallel to the flow direction) of 3 m. The airplane is flying at standard sea level with a velocity of 200 m/s. If the flow is considered to be completely laminar, calculate the boundary layer thickness at the trailing edge and the total skin friction drag. Assume that the wing is approximated by a flat plate. Assume incompressible flow.
Solution :
Given :
Rectangular wingspan
Length,L = 17.5 m
Chord, c = 3 m
Free stream velocity of flow, [tex]$V_{\infty}$[/tex] = 200 m/s
Given that the flow is laminar.
[tex]$Re_L=\frac{\rho V L}{\mu _{\infty}}$[/tex]
[tex]$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$[/tex]
[tex]$= 4.10 \times 10^7$[/tex]
So boundary layer thickness,
[tex]$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$[/tex]
[tex]$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.0024 m
The dynamic pressure, [tex]$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$[/tex]
[tex]$ =\frac{1}{2} \times 1.225 \times 200^2$[/tex]
[tex]$=2.45 \times 10^4 \ N/m^2$[/tex]
The skin friction drag co-efficient is given by
[tex]$C_f = \frac{1.328}{\sqrt{Re_L}}$[/tex]
[tex]$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.00021
[tex]$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$[/tex]
[tex]$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$[/tex]
= 270 N
Therefore the net drag = 270 x 2
= 540 N
When driving across Death Valley in the summertime, it is recommended that you release some air from your tires before making the crossing. Using the principles of Kinetic Molecular Theory (KMT), explain why it is a good idea to follow this recommendation.
According to the ideal gas law, pressure will rise as a gas's temperature rises. There is a limit to how much the tire can expand before the rubber gives in to the pressure build-up.
What the principles of Kinetic Molecular Theory?For every 10 degrees that the temperature drops, the inflation pressure in tires typically decreases by 1 to 2 psi. Moreover, as the tire pressure heats up during the first 15 to 20 minutes of driving, it will increase by one psi every five minutes.
The ideal gas law states that pressure will increase as a gas's temperature increases. Before the rubber gives in to the pressure build up, the tire can only expand so far.
Therefore, The pressure in your tires will increase due to the increased particle movement in hot air, which will cause the centre of the tread to bow out and wear out first. Increasing the demand for new tires.
Learn more about Molecular Theory here:
https://brainly.com/question/15013597
#SPJ5
The variable ______________ describes how quickly something moves.
it's up in Gogle trust me
man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the velocity of rain relative to the man will be?
Answer:
No answer
Explanation:
no explanation
The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.
Answer:
a) k = 701.8 N / m, b) m_{ast} = 61.1 kg, c) v ’= -1.3 10⁻⁴ m / s
Explanation:
a) For this exercise let's use the relationship of the angular velocity
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
k = w² m
the angular velocity is related to the period
w = 2π / T
we substitute
k = 4 π² [tex]\frac{m}{T^2}[/tex]
let's calculate
k = 4 π² 10 /0.75²
k = 701.8 N / m
b) now repeat the measurement with an astronaut on the chair
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where the mass Month the mass of the chair plus the mass of the astronaut
M = m + [tex]m_{ast}[/tex]
M = k / w²
w = 2π / T
let's calculate
w = 2π / 2
w = π rad / s
M = 701.8 /π²
M = 71,111 kg
now we use that
M = m + m_{ast}
m_{ast} = M - m
m_{ast} = 71.111 - 10.0
m_{ast} = 61.1 kg
c) if the astronaut's movement is simple harmonic
x = A cos wt
therefore the speed is
v = [tex]\frac{dx}{dt}[/tex]
v = -Aw sin wt
maximum speed is
v = - Aw
v = 0.100 π
v = 0.31416 m / s
we can suppose that the movement of the space station and the astronaut is equivalent to division of the same
initial instant. Before the move
p₀ = 0
final instant. When the astronaut is moving
p_f = M_station v’+ m_{ast} v
the moment is preserved
p₀ = pf
0 = M__{station} v ’+ m_{ast} v
v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]
we substitute
v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]
v ’= -1.3 10⁻⁴ m / s
the negative sign indicates that the station is moving in the opposite direction from the astronaut
What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds
pdf
Due date: February 22, 2021
10:00 AM EST
5: Holt SF 05Rev 43 - 10.0 pts possible
A 0.290 kg block on a vertical spring with a
spring constant of 4.65 x 103 N/m is pushed
downward, compressing the spring 0.0500 m.
When released, the block leaves the spring
and travels upward vertically.
The acceleration of gravity is 9.81 m/s.
How high does it rise above the point of
release?
Answer in units of m.
x x
A dropped ball gains speed because
its nature is to become closer to Earth,
its velocity changes.
a gravitational force acts on it
Of inertia
Answer:
3 and 3 and 3
Explanation:
I am sure Hope for brain list
An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.
The lever shown above can be used to move the
bowling ball off the shelf. Pushing down at what
point on the lever would require you to apply the
least amount of force to move the ball?
A. 1
B. 2
C. 3
D. 4
Answer:
C your answer would be C
Explanation:
It should be right
Two forces are exerted on an object in the vertical direction: a 20 N force downward and a 10 N force upward. The mass of the object is 25 kg. (1) What are some possibilities about the motion of this object? (2) Represent the motion of the object with a force diagram and a motion diagram.
Answer:
They are equal.
Explanation:
A storage tank has the shape of an inverted circular cone with height 12 m and base radius of 4 m. It is filled with water to a height of 10 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g
Answer:
Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2
Explanation:
Volume of Circular cone = V = (1/3)πr2h
where r is the radius in meters
and h is the height in meters
Substituting the given values in above equation, we get -
V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic meters.
The force required will be equal to the mass of water in the cone
[tex]= 167.47 * 1000[/tex]
= 167470 Kg
Weight = Mass * g
= 167470 * 10
= 1674700 Kgm/s^2
Cloudy days tend to have a greater range of temperatures than clear days. True or false?
Answer:
true
Explanation:
Answer:
true
Explanation:
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 3, 3, -3 > m/s, and the velocity of the other lump was < -4, 0, -4 > m/s. What is the velocity of the stuck-together lump just after the collision
Answer:
[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
Explanation:
[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]
[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]
m = Mass of each lump = [tex]30\ \text{g}[/tex]
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].