Explanation:
Given that,
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.
We need to find their speeds after 47.6 ns.
For electron,
The electric force is given by :
[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]
Let a be the acceleration of the electron. So,
F = ma
m is mass of electron
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]
Let v be the final velocity of the electron. So,
v = u +at
u = 0 (at rest)
So,
[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]
For proton,
Acceleration,
[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]
Now final velocity of the proton is given by :
[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]
Hence, this is the required solution.
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.3 m above the river, whereas the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 54.0 m wide. 1. How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? 2. What is the speed of the car just before it lands safely on the other side?
Answer:
Explanation:
1 ) Let the initial horizontal velocity of car be v .
For vertical displacement
vertical displacement h = 21.3 - 2.3 = 19 m
Time taken to fall by 19 m be t
19 = 1/2 x 9.8 t² ( initial downward velocity is zero )
t = 1.97 s
This is also the time taken to cover horizontal displacement of 54 m which is width of river .
horizontal speed v = 54 / 1.97 m /s
v = 27.41 m /s
2 )
At the time of landing on other side , car will have both vertical and horizontal speed .
vertical speed
v = u + gt
= 0 + 9.8 x 1.97 = 19.31 m /s
horizontal speed will remain same as the initial speed = 27.41 m /s
Resultant speed = √ ( 27.41² + 19.31² )
= √ ( 751.3 + 372.87)
= 33.52 m /s
An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement of a beam to which it is attached. (a)If an impact to the beam imparts a vibration at 4500 Hz, calculate the dynamic error and phase shift in the accelerometer output. (b)Calculateits resonance frequency.(c)What isthe maximumpossiblemagnitude ratio that the system can achieve
Answer:
A) i) Dynamic error ≈ 3.1%
ii) phase shift ≈ -12°
B) 79971.89 rad/s
Explanation:
Given data :
Damping ratio = 0.5
natural frequency = 18,000 Hz
a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz
i) Dynamic error
This can be calculated using magnitude ratio formula attached below is the solution
dynamic error ≈ 3.1%
ii) phase shift
This phase shift can be calculated using frequency dependent phase shift formula
phase shift ≈ -12°
B) Determine resonance frequency
Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s
C) The maximum magnitude ratio that the system can achieve
Explain why two electric charges of the same magnitude, when on a collision course with each other, won't actually collide
(serious answers only)
Answer:
Explanation:
When two charges of equal magnitude and sign approach each other, they interact through Coulomb's law
F = [tex]k \frac{q_ 1q_2}{z^2 }[/tex]k q1 q2 / r2
In you case the house are of equal magnitude and sign
q1 = q2 = q
F = k q2 / r2
Let's analyze this expression, the charge is repulsive on each charge, when they are on a collision course as they approach they feel an electric field opposite to their direction of movement, this field decreases its speed, the closer they get, the greater the repulsive force. , up to the point where this force is equal to or greater than the impulse, therefore the point where the velocity reaches zero, for this reason the particles do not actually touch
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.17 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 10.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
16.6 m
Explanation:
Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.
Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.
At catch up, D = d'
d + vt = v't
d = v't - vt
d = (v' - v)t
Substituting the values of the variables into d, we have
d = (2.75 m/s - 1.17 m/s)10.5 s
d = (1.58 m/s)10.5 s
d = 16.59 m
d ≅ 16.6 m
So, the other person was 16.6 m ahead of you when you started running.
Two bodies, A and B, have equal kinetic energies. The mass of A is nine times that of B. The ratio of the momentum of A to that of B is:_______.A. 1:9B. 1:3C. 1:1D.3:1E. 9:1
Answer:
D. 3 : 1
Explanation:
Let suppose that A and B are particles. From statement we know that [tex]K_{A} = K_{B }[/tex] (same kinetic energy) and [tex]m_{A} = 9\cdot m_{B}[/tex]. Then,
[tex]K_{A} = K_{B}[/tex]
[tex]\frac{1}{2}\cdot m_{A}\cdot v_{A}^{2} = \frac{1}{2}\cdot m_{B}\cdot v_{B}^{2}[/tex]
[tex]m_{A} \cdot v_{A}^{2} = m_{B}\cdot v_{B}^{2}[/tex]
[tex]\frac{v_{B}}{v_{A}} = \sqrt{\frac{m_{A}}{m_{B}} }[/tex]
[tex]\frac{v_{B}}{v_{A}} = 3[/tex]
And the ratio of the momentum of A to the momentum of B is:
[tex]r = \frac{m_{A}\cdot v_{A}}{m_{B}\cdot v_{B}}[/tex]
[tex]r = 9\times \frac{1}{3}[/tex]
[tex]r = 3[/tex]
Hence, the correct answer is D.
Use the worked example above to help you solve this problem. The amount of charge that passes through a filament of a certain lightbulb in 2.09 s is 1.56 C. (a) Find the current in the bulb. A (b) Find the number of electrons that pass through the filament in 5.24 s. electrons (c) If the current is supplied by a 12.0-V battery, what total energy is delivered to the lightbulb filament
Answer:
a) i = 0.746 A, b) # _electron = 2.44 10¹⁹ electrons, c) E = 1.87 10⁴ J
Explanation:
a) The definition of current is the charge per unit of time
i = Q / t
i = 1.56 / 2.09
i = 0.746 A
b) Let's look for the cargo in passing at this time
i = Q / t
Q = i t
Q = 0.746 5.24
Q = 3.904 C
an electron has a charge e = -1.6 10⁻¹⁹ C, let's use a direct proportions rule
# _electron = 3.904 C (1 electron / 1.6 10⁻¹⁹)
# _electron = 2.44 10¹⁹ electrons
the number of electrons has to be an integer
c) In this part you are asked to calculate the power
P = V i
P = 12 0.746
P = 8.952 W
P = E/t
E = P t
E = 8.952 2.09
E = 1.87 10⁴ J
The magnitude obtained when adding vector A (80 N at 20 deg) with vector B (40 N at
70 deg) is:
110.06 N
89.85 N
0 130.32 N
0 141.98 N
Answer:
110.06NExplanation:
The magnitude of the force is known as the resultant.
R = √Fx²+Fy²
Fx = 80cos 20 + 40cos70
Fx = 80(0.9397)+40(0.3420)
Fx = 75.176 + 13.68
Fx = 88.856N
Fy = 80sin 20 + 40sin70
Fy = 80(0.3420)+40(0.9397)
Fy = 27.36 + 37.588
Fy = 64.948N
R = √88.586²+64.948²
R = √7,847.48+4,218.24
R = √12,065.72
R = 109.5
R = 110N
Hence the magnitude of the forces is 110N
What is the density of a 36 g object with a volume of 15 cm3? (Density: D = )
0.42 g/cm3
0.54 g/cm3
2.4 g/cm3
5.4 g/cm3
Answer:
density = mass/volume
so . . .
density = (36 g)/(15 cm³) = 2.4 g/cm³
Explanation:
A 107 kg boat that is 7 m in length is initially 7.3 m from the pier. A 53 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. 7.3 m 7 m How far is the child from the pier when she reaches the far end of the boat
Answer:
11.98 m
Explanation:
Given that:
mass of the child [tex]m_c[/tex] = 53 kg
mass of the boat [tex]m_b[/tex] = 107 kg
[tex]\text{length of the boat L = 7 m}[/tex]
the distance of the boat from pies l = 7.3 m
initial momentum [tex]P_i = 0[/tex]
Final momentum [tex]P_f = mc \dfrac{L}{f}- (m_c +m_b) \dfrac{x}{l}[/tex]
where;
x = distance moved by boat towards left
t = time taken for the child to travel to the far end of the boat
[tex]P_i =P_f[/tex]
∴
[tex]m_c \dfrac{L}{t}=(m_c +m_b) \dfrac{x}{t}[/tex]
Then;
[tex]x = \dfrac{m_cL}{m_c+m_b}[/tex]
[tex]x = \dfrac{53 \times 7}{53+107}[/tex]
x = 2.32 m
The distance of the child from the pier is:
d = L +(l - x)
d = 7 m + ( 7.3 m - 2.32 m)
d = 7 m + 4.98 m
d = 11.98 m
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.
Answer:
15m/s
Explanation:
[tex]20m = \frac{1}{2} g {t}^{2} \\ t = 2s \\ t0 = 2s - 1s = 1s \\ vt + \frac{1}{2} g {t}^{2} = 20m \\ v \times 1s + \frac{1}{2} \times 10 \times 1 = 20 \\ v = 15[/tex]
A 40.0-kg packing case is initially at rest on the floor of a 1500-kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case
(a) when the truck accelerates at 2.20m/s22.20m/s 2 northward and
(b) when it accelerates at 3.40m/s23.40m/s 2 southward.
Answer:
Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.
The answer:
(a) 88N, northward.
(b) 78.4, southward.
Explanation:
(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)
Normal force = mass X acceleration due to gravity
Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case X acceleration due to gravity)
Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N
While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)
Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N
With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.
The answer is 88N, northward.
(b) Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.
Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136 N
We can conclude that the reaction force (136 N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.
This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:
kinetic force = (coefficient of kinetic friction) X (mass of packing case X acceleration due to gravity)
= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4N
hii guys!! ~ i've comeback again!! could you please help me with an assignment?? i have a couple of i know questions!! thank you for helping me, hehe.
tenise can run the 100 meter dash in 9.77 seconds. what is her speed??
Answer:
Speed = 10.24 m/s.
Explanation:
Given the following data;
Distance = 100m
Time = 9.77
To find her speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
Substituting into the equation, we have;
[tex]Speed = \frac{100}{9.77}[/tex]
Speed = 10.24 meter per seconds.
A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the
roof as shown in the figure below. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in
equilibrium. The mass of the sign is 27.5 kg, and the mass of the potted plant is 67.5 kg.
Plant
sale
today!
(a) Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the
potted plant.
N
(b) What is the maximum value of the static friction force if the coefficient of static friction between the pot and the
roof is 0.707?
N
Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.
Explanation:
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of a highway. Knowing that the speed of the bus was 80 mph as it begins to go up the hill and that the driver does not change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus up the hill when its speed decreased to 50 mph.
Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)
∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - [tex]a_{Net}[/tex] × t
∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex] = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
Pls help me I’m failing
Answer:
A is the answer okkkkkkkkkkkkkkkk
The Willis Tower in Chicago has an observation deck 412 m above ground.
How far can you see out over Lake Michigan from the observation deck?
Answer:
Check Newton's Rings:
d = height of air film
s = distance from center to ring being considered
R = radius of circle considered
The approximate formula is:
d = s^2 / (2 R) or s = (2 R d)^1/2
If we just use 4000 mi for R and 1/4 mi for d the height
we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi
is 250 000 miles from the earth to the moon" is a qualitative
Observation
TRUE
Or false
Answer:
True
Explanation:
Calculate the amount of heat given off by 640 g of water cooling from 76 °C to 28° C. Specific heat of water = 4.816 J/g C. Show your step by step process on how you have arrived at your answer. *
Answer:
47947.52 J.
Explanation:
From the question,
Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1
Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.
Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.
Substitute these values into equation 1 above
Q = 640×4.816(48)
Q = 147947.52 J.
Hence the amount of heat given off is 47947.52 J.
What two air masses creates hurricanes?
Answer:
The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.
Explanation:
Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.
Answer:
air
Explanation:
Please help! THIS IS A EASY ONE, HOPEFULLY.
Answer:
megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter
Explanation:
plz mark brainlest
Two loudspeakers, A and B are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. The frequency of the sound waves produced by the loudspeakers is 206 Hz. Consider a point P between the speakers and along the line connecting them, a distance x to the right of A. Both speakers emit sound waves that travel directly from the speaker to point P. For what values of x will:
a. destructive interference occur at P
b. constructive interference occur at P
Answer:
a) 0, ±1.65 b) ± 0.825m
Explanation:
This is a sound interference exercise, it can be described by the path difference between the two waves
for the case of constructive interference
Δr = 2n λ/ 2 n = 0, 1, 2 ...
for the case of destructive interference
Δr = (2n + 1) λ/ 2
the speed of sound is related to the wavelength
v = λ f
λ = v / f
λ= 340/206
λ = 1.65 m
let's set a reference system in the center between the two speakers
a) let's find the distances for constructive interference
Δr = 2n 1.65 / 2
Δr = 1.65 n
* the first interference occurs at n = 0
Δr = 0
therefore the interference in the center is maximum
* n = 1
Dr = 1.65 m
the second inference occurs at 1.65 m from the center, therefore there is a right wing and a left wing,
We do not have any more interference between the speakers because
n = 2 Δr = 3.3m this distance can be from the speaker
b) let's look for the destructive interference points
Δr = (2n + 1) 1.65 / 2
Δr = (2n + 1) 0.825
m = 0 Δr = 0.825m
m = 1 Δr = 2,475m
We can see that we only have the first destructive interference, one on each side.
what is a resultant vector
Answer:
A resultant vector is a combination or, in simpler words, can be defined as the sum of two or more vectors which has its own magnitude and direction.
reason why the center of gravity must not be at 50cm
Answer:
hope this helps
hope this is what u want
Consider the balloon and air inside the flask to be a closed system. Use the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Balloon will burst, due to more heat and in another flask air gets heated.
What is the first law of thermodynamics ?"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."
What is heat ?"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."
Know more about heat here
https://brainly.com/question/1429452
#SPJ3
Derase
An electric heater Consumes 1.8 MJ When connected to a 250V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply
Answer:
a. Power = 1000 Watts or 1 Kilowatts.
b. Current = 4 Amperes.
Explanation:
Given the following data;
Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules
Voltage = 250V
Time = 30 minutes to seconds = 30 * 60 = 1800 seconds
To find the power rating;
Power = energy/time
Substituting into the equation, we have;
Power = 1800000/1800
Power = 1000 Watts or 1 Kilowatts.
b. To find the current taken from the supply;
Power = current * voltage
1000 = current * 250
Current = 1000/250
Current = 4 Amperes.
pers
2. (a) Calculate the virtual depth of a black dot at the
bottom of a cubic block made of transparent glass with each
side 4 cm, while the refractive index of glass is 1.6.
Answer:
2.5 cm
Explanation:
Using the relation :
Refractive index = Real Depth / Apparent depth
Refractive index = 1.6
Real depth = 4cm
Virtual depth = apparent depth = x
1.6 = 4cm / x
1.6x = 4
x = 4 / 1.6
x = 2.5
Hence, virtual depth = 2.5cm
Clouds
Clouds have a big influence on weather. They are a necessary precursor of precipitation, although not all of them
produce precipitation. Clouds also prevent some solar radiation from reaching the ground and absorb some of the
heat that is re-radiated from the surface. As a result, cloudy days are likely to be cooler and cloudy nights warmer
than clear days and nights.
Water vapor condenses out of the air when the temperature reaches the dew point. Air may reach its dew point
when humidity increases or air temperature decreases. The latter commonly happens when warm, moist air rises.
For clouds to form, water vapor must condense around tiny particles called nuclei (singular, nucleus). A nucleus
might be a speck of dust or smoke, or it might be a salt crystal. The condensation of many water molecules around
a nucleus forms a tiny droplet of liquid water. If billions of these water droplets come together, they make a cloud.
Clouds are classified in several ways. The most common classification used today divides clouds into groups based on altitude
• Middle clouds form at middle altitudes and consist of ice crystals, water droplets, or both. Examples of middle
clouds include altocumulus and altostratus clouds.
• Low clouds form at low altitudes and consist entirely or mainly of water droplets. Examples of low clouds
include stratus, stratocumulus, and nimbostratus clouds.
• Vertical clouds grow upward and have their bases at low altitude and their tops at middle or high altitude.
They form when strong air currents carry warm air upward. Examples of vertical clouds include cumulus and
nimbocumulus clouds.
Questions
1. How do clouds influence weather?
2. Explain how clouds form.
3. Outline how clouds are classified by altitude
Answer:
clouds from when moist, warm rising air cools and expands in the atmosphere. The water vapor in the air condenses to form tiny water droplets which are the basis of clouds.
Explanation:
Find the least frequency of incident light that will knock electrons out of the surface of a metal with a work function of 3eV.
Answer:
The frequency must be: [tex]725\,\,10^{12} \,Hz[/tex]
Explanation:
If the work function of the metal ([tex]\phi[/tex]) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:
[tex]KE= h*f-\phi[/tex]
considering for the minimum KE = 0, and using the Plank constant h in eV s as: 4.14 * 10 ^(-15) eV s, to solve for the frequency:
[tex]h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz[/tex]
Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain
Answer:
Explanation:
Yes , it is possible .
When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .
Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .
V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .
A student carries a backpack for one mile. Another student carries the same backpack for two miles
Compared to the first student, how much work did the second student do?
Answer:
Compared to the first student, the second student did twice as much work as the first student.
Explanation:
The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance). The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).