Explanation:
The potential energy of two particles system will be given by the negative of work by gravitational force due to particle (1)as the second particle is brought from infinity : Gravitational potential energy Second particle is brought in the gravitation of first particle.
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]
calculate
q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
TWO forces, one of 12N and another or 24N
act on body in such a way that they make an angle of 90° with each other. Find the resaltant of two forces.
Answer:
26.833 N
Explanation:
The computation of the resaltant of two forces is shown below:
Given that
Force A = 12N
Force B = 24N
Based on the above information
Resultant R is
[tex]=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}[/tex]
=26.833 N
Help plsssssssssss I write it 100 time no one answer
Answer:
1.93×10²⁸ s
Explanation:
From the question given above, the following data were obtained:
Number of electron (e) = 2×10²⁴
Current (I) = 10 A
Time (t) =?
Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:
1 e = 96500 C
Therefore,
2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e
2×10²⁴ e = 1.93×10²⁹ C
Thus, 1.93×10²⁹ C of electricity is passing through the point.
Finally, we shall determine the time. This can be obtained as follow:
Current (I) = 10 A
Quantity of electricity = 1.93×10²⁹ C
Time (t) =?
Q = it
1.93×10²⁹ = 10 × t
Divide both side by 10
t = 1.93×10²⁹ / 10
t = 1.93×10²⁸ s
Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point
What would we need to do to make an electromagnet strong enough to move cars and trains
Answer:
The combined magnetic force of the magnetized wire coil and iron bar makes an electromagnet very strong. In fact, electromagnets are the strongest magnets made. An electromagnet is stronger if there are more turns in the coil of wire or there is more current flowing through it.
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
how many pennies can 4 folds of a paper hold?
walking dancing and even some household chores are?
Answer:
Actions
Explanation:
if u mean what are these called then it's actions
Answer: Regular Physical Activity
Explanation:
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.
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
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].
A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?
Answer:
is there a. b. c or d?
Explanation:
A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision?
Answer:
We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.
Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.
Explanation:
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 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
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.
The current flow in the light bulb is 0.5A
a.Calculate the amount of electric charge that flow through the bulb in 2 hour
b.If one election carries a
charge 1.6 x 10-14 c Find the number of election through the bulb in 2 hour?
Answer:
Explanation:
Given that,
The current in the light bulb, I = 0.5 A
(a) We know that,
Electric current = charge/time
or
Q = It
Put t = 2 hours = 7200 s
So,
Q = 0.5 × 7200
Q = 3600 C
(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]
Let there are n electrons flow through the bulb in 2 hours.
I = Q/t
Since, Q = ne
So,
I = ne/t
[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]
Hence, this is the required solution.
.................,,,,,,,,,,,
Answer:
B
Explanation:
Motion is movement, the teacher's movement is motion
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
Stored energy due to vertical position is known as
Elastic Potential energy
Vibrational energy
Kinetic energy
O Gravitational Potential energy
1
2
3
4
5
Answer: gravitational potential energy
Explanation:
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
Which describes the length of the power cord when Shanti gets off the train?
cannot be determined
less than 1.2 m
more than 1.2 m
equal to 1.2 m
Answer:
D. equal to 1.2
Explanation:
on edg
The length of the power cord will be equal to 1.2 m.
Describe about the length of power cord? The train is moving at a speed of 90 km /hr. Train was moving but the person in the train can be considered to be at rest. Shanti is the person travelling on the train. Her cord can be used only by her and the cord length of the phone will be 1.2 m.The length can be measured through the distance.The unit of length is meter.As we know the concept of motion and rest, there only the train in motion, shanti was at rest and shanti's power cord were also in the rest. Power cord length will be determined only at the time of manufacturing.If the power cord length to be change then the crimping process.So, the length will not change suddenly.
The length of the power cord when shanti gets off the train is equal to 1.2 m.
The Correct answer is Option D.
Learn more about motion and rest,
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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
Which is an example of kinetic energy?
A. The energy stored in
ethanol
B. A ball sitting at the top of a ramp
C. A compressed spring
D. A hockey puck sliding across ice
D. A hockey puck sliding across ice
4 Two friction disks A and B are to be brought into contact withoutslipping when the angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at time t = 0 and is given a constantangular acceleration with a magnitude α. Disk B starts from rest attime t = 2 s and is given a constant clockwise angular acceleration,also with a magnitude α. Determine (a) the required angular acceleration magnitude α, (b) the time at which the contact occurs
This question is incomplete, the missing image is uploaded along this answer below;
Answer:
a) the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) the time at which the contact occur is 8 seconds
Explanation:
Given the data in the question;
first we convert the given angular velocity to rad/s
angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s
so
ωA = 8π rad/s
next we determine angular acceleration at point A; so
ωA = at
8π rad/s = at -------let this be equation
thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.
Next we determine the velocity of point C;
Vc = rA × ωA
where Vc is velocity at point C, rA is radius of A ( 150/1000)m, { from the diagram }
so we substitute
Vc = 0.15m × 8π
Vc = 1.2π m/s
for angular velocity at point B;
Vc = rB × ωB
where rB is the radius of B ( 200/1000)m
we substitute
1.2π = 0.2 × ωB
ωB = 1.2π / 0.2
ωB = 6π rad/s
Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.
Now,
a) Determine the required angular acceleration magnitude α
we find the the angular acceleration of disk B after 2 seconds, using the expression;
ωB = at
where angular acceleration is a and t is time ( t - 2)
we substitute
ωB = at
6π = a( t - 2) -------- let this be equation 2
now, lets substract equation 1 form equation 2
(6π = a( t - 2)) - (8π = at)
(6π = at - 2a) - ( 8π = at)
-2π = 0 + -2a
2π = 2a
a = 2π/2
a = π rad/s² or 3.14 rad/s²
Therefore, the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) determine the time at which the contact occurs;
from equation 1
8π = at
we substitute in the value of a
8π = π × t
t = 8π / π
t = 8 seconds
Therefore, the time at which the contact occur is 8 seconds
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
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Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.
Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
A baseball is thrown horizontally from a cliff at 30 m/s and lands 7 seconds after the baseball was thrown. Calculate the horizontal AND vertical distance.
Answer:
The horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively.
Explanation:
Using the equation of the displacement in the x-direction, we have:
(let's recall we have a constant velocity in this direction)
[tex]x=v_{ix}t[/tex]
Where:
v(ix) is the initil velocity in the x direction (v(ix) = 30 m/s)t is the time (t = 7 s)[tex]x=30(7)[/tex]
[tex]x=210\: m[/tex]
Now, we need to use the equation of the displacement in the y-direction to find the vertical distance. Here we have an acceleration (g)
[tex]y=v_{iy}t-\frac{1}{2}gt^2[/tex]
Where:
v(iy) is the initial velocity at the y-direction. In this case, it will be 0t is the timeg is the acceleration of gravity (g=9.81 m/s²)Then, the vertical position at 7 s is:
[tex]y=-\frac{1}{2}(9.81)(7)^2[/tex]
[tex]y=-240.35\: m[/tex]
Therefore, the horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively. The minus sign means the negative value in the y-direction.
I hope it helps you!
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(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:
.
What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds
A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?
The impulse exerted on the beach ball is 2.75 kgm/s.
The impulse exerted on the exercise ball is - 2.75 kgm/s.
What is impulse?
This is the force applied to an object that acts over a period of time.
The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;
J = ΔP
J = m(v - u)
J = 0.25(7 - (-4))
J = 0.25(7 + 4)
J = 2.75 kgm/s
The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.
Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.
Learn more about impulse here: https://brainly.com/question/25700778
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