A 4.0-g bead carries a charge of 20 μC. The bead is accelerated from rest through a potential difference V, and afterward the bead is moving at 3.0 m/s. What is the magnitude of the potential difference V? *
1 point
900 V
400 V
200 V
400 kV
Answer:
The magnitude of the potential difference is 900 V.
Explanation:
Given;
mass of the bead, m = 4.0 g = 0.004 kg
charge of the bead, Q = 20 μC = 20 x 10⁻⁶ C
final velocity of the bead, v = 3 m/s
What is the magnitude of the potential difference V?
Apply the principle of conservation of energy;
The electric potential energy at the beginning is equal to kinetic energy of the bead at the end of the journey.
qV = ¹/₂mv²
[tex]V = \frac{mv^2}{2q} \\\\V = \frac{0.004 \ \times \ (3)^2}{2(20 \times 10^{-6})}\\\\V = 900 \ V[/tex]
Therefore, the magnitude of the potential difference is 900 V.
The magnitude of the potential difference (V) is equal to 900 Volts.
Given the following data:
Mass of bead = 4.0 g to kg = 0.004 kgCharge of bead = 20 μC = [tex]20 \times 10^{-6} \;C[/tex]Final velocity of bead = 3 m/sTo determine the magnitude of the potential difference (V):
How to calculate the potential difference (V).We would apply the law of conservation of energy, which states that the electric potential energy possessed by the bead at the beginning is equal to the kinetic energy possessed by the bead at the end of the journey:
[tex]qV = \frac{1}{2} mv^2\\\\V = \frac{\frac{1}{2} mv^2}{q} \\\\V = \frac{\frac{1}{2} \times 0.004 \times 3.0^2}{20 \times 10^{-6}} \\\\V = \frac{ 0.002 \times 9}{20 \times 10^{-6}}[/tex]
V = 900 V.
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How do you classify flammable liquid to gas and solid?
Answer:
A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.Add to that a gaggle of state, regional, and local authorities; and we feel compelled to begin this blog entry with our favorite caveat: get expert advice before deciding what to do with that rusting drum of stale gasoline out back.Explanation:
I hope it help you;)
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?
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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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]
Why are we seeing extremely old light from Canopus instead of light in real-time?
Answer:
Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.
At what tempreture will the of and oC
be the same
Answer:
-40 degreesTo find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
Explanation:
Hope it is helpful.....
Answer:
To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.
A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
0.248
Explanation:
Initial speed u = 13.6
Final speed v = 0
Distance s = 38.1
We have umg = ma
We make u subject of the formula
u = a/g
V² = u² + 2as
a = v²-u²/2s
We substitute the values into the above
a = 0-(13.6)²/2*38.1
a = 184.96/76.2
a = 2.427m/sec
Remember that
u = a/g
u = 2.427/9.8
= 0.2476
This is approximately 0.248
This is the minimum coefficient of friction required to keep the crate from sliding.
a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?
Answer:
2.5 m/s^2
Explanation:
First, convert 90 km/hr into m/s:
90/3.6 = 25 m/s
vf = final velocity = 25 m/s
vi = initial velocity = 0 m/s
t = time = 10 seconds
a = acceleration, unknown
Then, find a using the following equation:
(vf - vi)/t = a
(25 m/s)/10 s = 2.5 m/s^2
a = 2.5 m/s^2
Hope this helps!! :)
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.
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attached to hanging blocks. The block attached to string 1 has a mass of 20 kg and the block attached to string 2 has a mass of M. Listeners hear a beat frequency of 2 Hz when string 1 is excited at its fundamental frequency and string 2 is excited at its third harmonic. What is one possible value for mass M
Answer:
2.18 kg
Explanation:
The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.
For string 1, its fundamental frequency f is when n = 1. So,
f = 1/2L√(T/μ) = 1/2L√(mg/μ)
Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f = 1/2L√(mg/μ)
f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)
f = 1/1 m√(196 kgm/s²/0.005 kg/m)
f = 1/1 m√(39200 m²/s²)
f = 1/1 m × 197.99 m/s
f = 197.99 /s
f = 197.99 Hz
f ≅ 198 Hz
For string 2, at its third harmonic frequency f' is when n = 3. So,
f' = 3/2L√(T/μ) = 3/2L√(mg/μ)
Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m
substituting the values of the variables into f, we have
f' = 3/2L√(Mg/μ)
f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)
f' = 3/1 m√(M1960 m²/s²kg)
f' = 3/1 m√M√(1960 m²/s²kg)
f' = 3/1 m √M × 44.27 m/s√kg
f' = 132.81√M/s√kg
f' = 132.81√M Hz/√kg
Since the frequency of the beat heard is 2 Hz,
f - f' = 2 Hz
So, 198 Hz - 132.81√M Hz/√kg = 2 Hz
132.81√M Hz/√kg = 198 Hz - 2 Hz
132.81√M Hz/√kg = 196 Hz
√M Hz/√kg = 196 Hz/138.81 Hz
√M/√kg = 1.476
squaring both sides,
[√M/√kg] = (1.476)²
M/kg = 2.178
M = 2.178 kg
M ≅ 2.18 kg
For a given substance, the molecules
move fastest when the substance is
Answer:GAS
Explanation:
what is a vector quantity?
Answer:
A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude
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
Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.30 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.30 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant.
Required:
a. Calculate the time of Blythe's run.
b. Calculate the time of Geoff s run.
Answer:
Explanation:
1.00 km = 1000 m .
Blythe's run : -------
Time taken to run 600 m speed
= distance / speed
T₁= 600 / 4.10 = 146.34 s
Next time T₂ = 1 min = 60 s
acceleration of Blythe
a = (7.30 - 4.10) / 60 = .053 m /s²
displacement during acceleration
= ut + 1/2 at²
= 4.10 x 60 + .5 x .053 x 60²
= 246 + 95.4
= 341.4 m
Rest of the distance to be covered = 1000 - ( 600 + 341.4 )
= 58.9 m
Time taken to cover this distance
T₃= 58.9 / 7.3 = 8.06 s
Total time = T₁ + T₂ + T₃ = 214.4 s
Geoff s run : ---------
initial acceleration during first 3 min
= (8.3 - 0 ) / (3 x 60 )
= .046 m /s²
displacement
s = ut + 1/2 a t²
= 0 + .5 x .046 x ( 3 x 60 )²
= 745.2 m
Rest of the distance of race
= 1000 - 745.2 = 254.8 m
This distance is covered at speed of 8.3 m/s
time taken to cover this distance
T₂ = 254.8 / 8.3
= 30.7 s
Total time taken to complete the race
= 180 + 30.7
= 210.7 s .
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
Four-wheel drive trucks do not stop better on icy
roads than a car. Is what law of motion (Newton's laws)
please answer correctly
Answer:
616000 J.
Explanation:
Bi. Determination of the work done.
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:
Workdone = Force × distance
Wd = F × s
With the above formula, we can obtain the Workdone as follow:
Force (F) = 220 N
Distance (s) = 2800 m
Workdone (Wd) =?
Wd = F × s
Wd = 220 × 2800
Wd = 616000 J.
A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?
AnswerHM???
Explanation:
I dONT KNOW
Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answer: it is thin near the center and thick at the edges
Explanation: took the test on Plato :)
Plzz help me with this
Answer: d
Explanation:
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]
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.
Which of the following hydrocarbons are SATURATED hydrocarbons?
I. alkanes II. alkenes III. alkynes IV. cycloalkanes
A. I and IV
B. II and III
C. I and III
D. II and IV
Answer:
i think c
Explanation:
If a satellite is orbiting the Earth in elliptical motion, then it will move _______________ (slowest, fastest) when its closest to the Earth. While moving towards the Earth (along the path from D to A) there is a component of force in the __________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow down, speed up). While moving away from the Earth (along the path from A to D) there is a component of force in the _________________ (same, opposite) direction as the motion; this causes the satellite to ___________________ (slow
Answer:fastest,same,slow down,opposite,slow
Explanation:
A satellite move fastest when its closest to the Earth. The other correct options are same direction, speed up, opposite direction and slow.
Velocity of a satellite around the planet.If a satellite is orbiting the Earth in elliptical motion, then it will move fastest when its closest to the Earth (based on Kepler's, law).
While moving towards the Earth (along the path from D to A) there is a component of force in the same direction as the motion; this causes the satellite to speed up.
While moving away from the Earth (along the path from A to D) there is a component of force in the opposite direction as the motion; this causes the satellite to slow.
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#SPJ2
Plzz answer this question correctly
Answer:
by reducing friction.....
what is the mathematical formula associated with newton's 2nd law of motion?
Answer:
F= m x a
Explanation:
Force (f) = mass (m) x acceleration (a)
[RM.03H]Which of these is the most likely impact of extensive mining of uranium to produce energy?
land becomes unfit for food production
rainfall decreases because of harmful gases
greenhouse gases are absorbed by the mineral
radiations are better absorbed by the atmosphere
Answer:
land becomes unfit for food production
A flat screen tv uses 120 watts. How much energy is used up if it is left on for 15 min?
A.) 4j
B.) 15j
C.) 0.67j
D.) 108,000j
Answer:
d
Explanation:
An object initially traveling in a straight line with
a speed of 5.0 meters per second is accelerated
at 2.0 meters per second squared for 4.0 seconds.
The total distance traveled by the object in the
4.0 seconds is
Answer:
We conclude that the total distance traveled by the object in the 4 seconds is 36 m.
Explanation:
Given
Initial velocity u = 5.0 m/sAcceleration a = 2.0 m/s²Time t = 4 sTo determine
The total distance traveled by the object in the 4.0 seconds is
Important Tip:
We can determine the total distance traveled by the object in the 4.0 seconds by using the equation of motion such as
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = distanceu = initial velocitya = accelerationt = timesubstituting u = 5.0, a = 2, and t = 4 in the formula
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\left(5\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(4\right)^2[/tex]
[tex]s=20+16[/tex]
[tex]s=36[/tex] m
Therefore, we conclude that the total distance traveled by the object in the 4 seconds is 36 m.
The total distance traveled by the object is 36 meters.
Given the following data:
Initial velocity = 5.0 m/s Acceleration = 2.0 [tex]m/s^2[/tex] Time = 4.0 seconds.To find the total distance traveled by the object, we we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2}at^2[/tex]
Where:
S is the total distance traveled.
u is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting the values into the formula, we have;
[tex]S = 5(4) + \frac{1}{2}(2)(4^2)\\\\S = 20 + 1(16)[/tex]
Total distance, S = 36 meters.
Therefore, the total distance traveled by the object is 36 meters.
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Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Answer:
Explanation:
Total time taken = 0.85 s .
Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85
5.43 / 343 + time taken by coin to fall by 5.43 m = .85
time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s
Let the initial velocity of throw of coin = u
displacement of coin s = 5.43 m
time take to fall t = .834 s
s = ut + 1/2 gt²
5.43 = u x .834 + .5 x 9.8 x .834²
5.43 = u x .834 + 3.41
u x .834 = 2.02
u = 2.42 m /s .