Answer:
v₀ₓ = 24.24 m / s
Explanation:
This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.
Y axis
initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
when it reaches the ground its height is zero and the initial height is y₀=1.2m
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2 y_o/g}[/tex]
t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]
t = 0.495 s
X axis
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 12 / 0.495
v₀ₓ = 24.24 m / s
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate
Complete Question
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate(and 2.00 cm from the other) is 450 V?
Answer:
[tex]V'=562.5v[/tex]
Explanation:
From the question we are told that:
Separation distance [tex]d=10cm[/tex]
Voltage at 8cm [tex]V_8=450v[/tex]
Generally the equation for Voltage is mathematically given by
[tex]|V|=|E.d|[/tex]
Where
E=electric field
Therefore
At [tex]d=0.8[/tex]
[tex](450-0)V=E*(0.08m)[/tex]
[tex]E=\frac{450}{0.08}[/tex]
[tex]E=5625v/m[/tex]
Therefore
At [tex]d=10[/tex]
[tex]V'=Ed[/tex]
[tex]V'=5625*0.1m[/tex]
[tex]V'=562.5v[/tex]
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds
A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. The loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.
Answer:
the correct answer is B
Explanation:
Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved
instantly starts. Before the crash
p₀ = M v +0
final instant. After the crash
m_f = (M + M) v_f
the moment is preserved
p₀ = p_f
M v = 2 M v_f
v_f = v / 2
let's look for kinetic energy
before the crash
K₀ = ½ M v²
after the crash
K_f = ½ 2M (v_f)²
K_f = ½ 2M (v/2)²
K_f = (½ M v²) ½
K_f = K₀ / 2
therefore the correct answer is B
A certain electric stove has a 16 Ω heating element. The current going through the element is 15 A. Calculate the voltage across the element.
The voltage across the element is = 240 V
I hope you understand....
Mark me as brainliest....
Thanks...
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car? (Assume the bottom of the cliff is zero)
Answer:
Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.
modify other clauses and phrases in the sentence
refer to other sentences in the passage
make it hard to determine where the sentence ends
change the intended meaning of the sentence
Explanation:
In a 2-dimensional Cartesian coordinate system the y-component of a given vector is equal to that vector's magnitude multiplied by which trigonometric function, with respect to the angle between vector and y-axis?
a. sine
b. cosine
c. tangent
d. cotangent
Answer:
Option b, cosine.
Explanation:
Below you can see an image that illustrates this situation.
Remember that for a triangle rectangle with a given angle θ, we have:
Cos(θ) = (adjacent cathetus)/(hypotenuse)
In the image, you can see a vector of magnitude M, and the angle θ defined between the vector and the positive y-axis.
In this case, the y-component is the adjacent cathetus and the hypotenuse is the magnitude of the vector.
Then we will have:
Cos(θ) = (adjacent cathetus)/(hypotenuse) = y/M
solving that for y, we get:
y = Cos(θ)*M
Then the y-component is the vector's magnitude multiplied by the cosine of the angle between the vector and the y-axis.
The correct option is b.
Answer:
(b) cosine
Explanation:
In a 2-dimensional Cartesian coordinate system, a vector has a x-component and/or a y-component. To get these components, the magnitude of the vector is resolved with respect to the x-axis and the y-axis by multiplying it (the magnitude) by some trigonometric function with respect to the angle between the vector and the x or y axis.
For example, given a vector A of magnitude A which makes an angle α with the x-axis and an angle β with the y-axis, the x and y components of the vector A can be found as follows;
i. x-component is given by [tex]A_{x}[/tex]
[tex]A_{x}[/tex] = A cos α (with respect to the angle between A and the x-axis) or
[tex]A_{x}[/tex] = A sin β (with respect to the angle between A and the y-axis)
ii. y-component is given by [tex]A_{y}[/tex]
[tex]A_{y}[/tex] = A sin α (with respect to the angle between A and the x-axis) or
[tex]A_{y}[/tex] = A cos β (with respect to the angle between A and the y-axis)
Therefore, the y-component of a vector in a 2-dimensional Cartesian coordinate is given by the product of the magnitude of the vector and the cosine of the angle between the vector and the y-axis.
Khối lượng ban đầu của mặt trời m⊙ = 2 × 1030 kg trong đó có 71 % là
hydrogen. Trong 5×10^9 năm đầu tiên, mặt trời phát ra năng lượng với công suất
3.86 × 10^26 W, nhờ phản ứng
4p → 4+2He + 2e
+ 2νe + 26 MeV
trong đó bốn proton tổng hợp thành hạt α và tỏa ra năng lượng 26 MeV. Hãy
xác định
a) số hạt proton của mặt trời tại thời điểm ban đầu.
b) số hạt proton của mặt trời tham gia phản ứng trong 1 năm.
c) số hạt proton còn lại trong mặt trời sau 5 × 10^9 năm.
d) thời gian để tiêu thụ hết 10 % số hạt proton còn lại của mặt trời
Answer:
jsgssbvwsvs
Explanation:
ifmd understand
What is a reasonable measurement for the distance to Neptune?
30 light years
30 kilometers
30 parsecs
30 Astronomical Units
Answer:
30 kilometers is a reasonable measurement
Topic: Physical and Chemical Changes
Subject: Science
Grade: 5th
Question: Why am I growing?
(Please give this answer related to Physical and Chemical Changes.
Answer: Why am I growing is a chemical change.
Explanation:
Changes that can be reversed and does not affect the composition of a substance are called physical changes.
For example, change in state of a substance like ice converting into water is a physical change.
Changes that cannot be reversed and affect the chemical composition of a substance are called chemical changes.
For example, a child growing is an irreversible change and hence, it is a chemical change.
Thus, we can conclude that why am I growing is a chemical change.
A student must use an object attached to a string to graphically determine the gravitational field strength near Earth's surface. The student attaches the free end of the string to the ceiling and pulls the object-string system so that the string makes an angle of 5 degrees from the object's vertical hanging position. The student then releases the object from rest and uses a stopwatch to measure the time it takes for the object to make one complete oscillation. Which of the following is the next step that will allow the student to determine the gravitational field strength?
А) Repeat the experiment by adding additional mass to the object for multiple trials
B) Repeat the experiment by changing the length of the string for multiple trials
C) Repeat the experiment by changing the angle that the string makes with the object's vertical hanging position
D) Repeat the experiment by measuring the time it takes to make two oscillations, three oscillations, and additional oscillations for multiple trials
Answer:
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
Explanation:
The student is reacting a simple pendulum experiment where he can determine the value of the relationship of gravity with the expression
T = 2π [tex]\sqrt{\frac{L}{g} }[/tex]
let's analyze each statement
A) False. The mass is not a paramer of the period expression
B) True. By changing the length get a different period and with a graph you can find the best value of the gravity pull
C) False. The angle while it is small does not influence the period
D) True. By changing the number of oscillations the period does not change, so you can get the value of the pull of gravity.
We can see that the expressions B and d are true, the most exact value is obtained using procedure B since the graphs allow to reduce the errors
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = [tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}[/tex]
= [tex]4.59\times 10^{-19} \ J[/tex]
or,
= [tex]\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
= [tex]2.87 \ eV[/tex]
(b)
As we know,
⇒ [tex]Vq=\frac{hc}{\lambda}-\Phi_0[/tex]
By substituting the values, we get
⇒ [tex]1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0[/tex]
⇒ [tex]\Phi_0=2.3\times 10^{-19} \ J[/tex]
or,
⇒ [tex]=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
⇒ [tex]=1.4375 \ eV[/tex]
Suppose there are 3 molecules in a container. If each molecule has a 1-in-2 chance of being in the left half of the container, what is the probability that there are exactly 2 molecules in the left half of the container?
Answer:
Total probability = 3/8
Explanation:
Below is the calculation:
Number of molecules in the container = 3
The probability of one molecule in the left half, P = 3 / 2 = 1.5 or 1/2
The probability of second molecule in the left half, P1 = (3/4)
Total probability = P x P1
Total probability = (1/2) x (3/4)
Total probability = 3/8
The lumberjack pulls on the sled with 40 N at an angle of 30 degrees, pulling so the sled moves at a constant velocity. 1) What is the x component of the applied force? 2) What is the y component of the applied force? 3) If the loaded sled has a mass of 65 kg, what is the magnitude of the force of gravity? 4) What is the magnitude of the normal force acting on the sled? 5) What is the coefficient of friction between the snow and the sled?
1) (40 N) cos(30°) ≈ 34.6 N
2) (40 N) sin(30°) = 20 N
3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N
4) The net force on the sled acting in the vertical direction is made up of
• the sled's weight, 585 N, pointing downward
• the vertical component of the applied force, 20 N, pointing upward
• the normal force, with magnitude n, also pointing upward
The sled does not move up or down, so by Newton's second law,
∑ F = n + 20 N - 585 N = 0 ==> n = 565 N
5) The net force in the horizontal direction consists of
• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)
• kinetic friction, with magnitude f, pointing in the opposite and negative direction
By Newton's second law,
∑ F = 34.6 N - f = 0 ==> f ≈ 34.6 N
Now if µ is the coefficient of kinetic friction, then
f = µn ==> µ = f/n = (34.6 N) / (565 N) ≈ 0.0613
The component of the force is the effective part of that force in that direction.
What is the component of a force?The component of the force is the effective part of that force in that direction.
1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N
2) The vertical component of the force = 40 N sin 30 degrees = 20 N
3) The magnitude of the gravitational force = mg cos 30 degrees = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N
4) The normal force = 551.7 N
5) The coefficient of friction = F/R = 40 N /551.7 N = 0.07
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Azam had a metal pencil box and a wooden block on his table. When he touched them, the pencil box felt colder compared to the wooden block. When he measured their temperatures using a thermometer, he found that both the objects were at the same temperatures. What could be the reason for Azam to 'feel' that the pencil box was cooler? * (1 Point) Air around the pencil box was at a lower temperature which made the pencil box colder Wood produces heat and so, must have been at a higher temperature than the pencil box As metal is a better conductor of heat, it drew away heat from his hand faster than the wooden block. Metals are always at a lower temperature than other substances and so, there has been an error in measurement
The third choice is the correct explanation. Don't make me type it all out on my phone.
Use a projectile motion kinematic equation for the vertical motion to find the time t that the ball is in the air, from when it leaves the track until it strikes the floor.
Answer:
The time of flight is [tex]T = \frac{2 u sin A}{g}[/tex].
Explanation:
Let the initial velocity is u and the angle of projection is A.
Use first equation of motion for vertical motion
Let the time to reach the maximum height is t.
[tex]v = u - gt\\\\0 = u sin A - gt \\\\t = \frac{ u sin A}{g}[/tex]
Total time of flight is
T =2 t
[tex]T = \frac{2 u sin A}{g}[/tex]
A speedometer in a car gives the car’s speed at that given moment, or the?
A. General speed
B. Instantaneous speed
C. Average speed
D. Constant speed
It’s not C or D!
Answer:
a because it is at a given moment
Explanation:
did u
What actually heats up the atmosphere?
Answer:
The heat source for our planet is the sun. Energy from the sun is transferred through space and through the earth's atmosphere to the earth's surface. Since this energy warms the earth's surface and atmosphere, some of it is or becomes heat energy.
Determine the values of m and n when the following average magnetic field strength of the Earth is written in scientific notation: 0.0000451 T. Enter m and n, separated by commas.
Answer:
B = 4.51×10⁻⁵ T
Explanation:
Given that,
The average magnetic field strength of the Earth is 0.0000451 T.
We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :
N=a×bⁿ
Where
n is any integer and a is a real no
So,
0.0000451 = 4.51×10⁻⁵ T
So, the required answer is equal to 4.51×10⁻⁵ T.
What is the torque in ( lbs-ft ) of a man pushing on a wrench with 65 lbs of force 8 unches from the nut / bolt he is trying to turn?
Explanation:
The torque [tex]\tau[/tex] is given by
[tex]\tau=Fd = (65\:\text{lbs})(\frac{8}{12}\:\text{ft}) = 43.3\:\text{lbs-ft}[/tex]
A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.
Answer:
below
Explanation:
Part A) This engine works per cycle is 254.9 J.
Part B) The thermal efficiency is 23.42%
What is the thermal efficiency?The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.
For isothermal expansion, work done is
W₁ =nRT₁ x ln(V₂/V₁)
W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)
W₁ = 339.8 J =Q₁
For isochoric cooling ,
W₂ =0
Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)
Q₂ = -748.3 J
For isothermal compression,
W₃ =nRT₂ ln (V₄/V₃)
W₃ = 0.1 x 8.314 x 200 x ln(3000/5000)
W₃ = -84.9J
For isochoric heating
W₄ =0
Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)
Q₄ = -748.3 J
Total work done in all the process W = W₁ +W₂ +W₃ +W₄
W =254.9 J
Thus, the work done is 254.9 J
Thermal efficiency = Work done/Heat taken
η = W/ Q₁ +Q₄
η = [254.9 / 339.8 +748.3 ] x 100 %
η = 0.2342 x 100 %
η = 23.42%
Thus, the thermal efficiency is 23.42%
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'
Calculate the terminal velocity of a rain drop of radius 0.12cm
Explanation:
Given that,
The radius of rain drop, r = 0.12 cm = 0.0012 m
The viscocity of air is, [tex]\eta=18\times 10^{-5}\ poise[/tex]
Let the viscous force is, [tex]F = 0.010173\ N[/tex]
The viscous force is given by :
[tex]F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}[/tex]
Put all the values,
[tex]v=\dfrac{0.010173}{6\pi 18\times 10^{-5}\times 0.0012 }\\\\v=2498.58\ m/s[/tex]
40 ohms
1.2 A
40 ohms
12 V
Calculate the total energy developed in 5
minutes by the system above.
Answer:
17280 J and 1080 J
Explanation:
Given :
R= 40 ohm
I=1.2A
t= 5 min=60×5=300 sec
Now,
Total energy can be calculated as:
[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]
Now,
V=12V
R=40 Ohm
[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]
Total energy is 17280 J and 1080 J
two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times the inertia of the other .when the blocks are released ,the spring pushes them away from each other .what is the ratio of their kinetic energies after the release?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] for both blocks is zero. Let [tex]m_1= m[/tex] and [tex]m_2 =3m[/tex]. So using the conservation law of linear momentum, we can write
[tex]0 = m_1v_1 - m_2v_2[/tex]
or
[tex]v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1[/tex]
The final kinetic energies for the two masses are
[tex]KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2[/tex]
[tex]KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)[/tex]
Therefore, the ratio of their kinetic energies is
[tex]\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}[/tex]
It takes a minimum distance of 98.26 m to stop a car moving at 17.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
x_f = 212.5m
Explanation:
t = (x_f-x_0)/(.5*(v_f-v_0))
t = (98.26m-0m)/(.5(0m/s-17m/s))
t = 11.56s
a = (v_f-v_0)/t
a = (0m/s-17m/s)/11.56s
a = -1.47m/s²
t = (v_f-v_0)/a
t = (0m/s-25m/s)/-1.47m/s²
t = 17s
x_f = x_0+(.5*(v_f-v_0))*t
x_f = 0m+(.5*(0m/s-25m/s))*17s
x_f = 212.5m
A pitcher throws a curveball that reaches the catcher in 0.61 s. The ball curves because it is spinning at an average angular velocity of 350 rev/min (assumed constant) on its way to the catcher's mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher
Answer:
22.36 rad
Explanation:
Applying,
ω = θ/t.............. Equation 1
Where ω = angular velocity, θ = angular displacement of the baseball, t = time
make θ the subject of the equation
θ = ωt............... Equation 2
From the question,
Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s
Substitute these values into equation 1
θ = 0.61(36.652)
θ = 22.36 rad
Hence the angular displacement of the baseball is 22.36 rad
A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficients of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
12) A negatively-charged balloon touching a wooden wall
A) pulls positive charge on the wall surface toward it.
B) pushes negative charge in the wall away from it.
C) polarizes molecules in the wall.
D) all of the above.
Answer:
D) all of the above.
Explanation:
First polarises it, cahrging and discharging occurs at once.
A negatively charged balloon touching a wooden wall then from the given options option D is correct which is all of the above.
What is a charge?Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons, respectively).
Contrary charges attract one another, while like charges repel one another. A neutral object is one that carries no net charge. Classical electrodynamics, the name given to an early understanding of how charged particles interact, is still accurate for issues that do not call for taking into account quantum phenomena.
In the first step it polarizes molecules in the wall, then charging and discharging in the wall will take place at once.
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New alleles arising from mutations in a population will
A homeowner has a new oil furnace which has an efficiency of 60%. For every 100 barrels of oil used to heat his house, how much (in barrels of oil) goes up the chimney as waste heat?
Answer:
below
Explanation:
