Answer:
98.11 I think
Explanation:
I really hope this helps have a wonderful day
Mischievous Joey likes to play with his family's lazy susan (this drives Mom crazy because it is an antique). He puts the salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 79 grams. She also measures the radius of the turntable to be 0.23 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.7222 m/s.
Required:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
Answer:
0.179 N
Explanation:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.
So, F = mv²/r where m = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m
So, substituting the values of the variables into the equation, we have
F = mv²/r
F = 0.079 kg (0.7222 m/s)²/0.23 m
F = 0.0412 kgm/s² ÷ 0.23 m
F = 0.179 kgm/s²
F = 0.179 N
2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.
Answer:
3.277 m
Explanation:
Given :
Maximum Height (Hmax) = (u²sin²θ) / 2g
Xv = Xh + Uv * t + 0.5gt²
Xv and Xh are vertical and horizontal distances
-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2
-1 = 0 + 0.903Uv - 11.025
-1 + 11.025 = 0.903Uv
10.025 = 0.903Uv
Uv = 10.025 / 0.903
Uv = 11.10 m/s
Hmax = 1 + (u²sin²θ) / 2g
= (11.10^2 * (sin37)^2) / 2*9.8
= 44.624360 / 19.6
= 2.277
Hmax = 1 + 2.277
Hmax = 3.277 m
An iron block of 12 kg undergoes a process during which there is a heat gain from the block at 2 kJ/kg, an elevation increase of 32 m, and a decrease in velocity from 40 m/s to 7 m/s. During the process, which also involves work transfer, the internal energy of the block increases by 70 kJ. Suppose the total energy of the system remains constant. Determine the work transfer during the process in kJ and indicate whether the work is done on/by the system.
Answer:
Explanation:
Total heat gain by the block ΔQ = 2 x 12 kJ = 24 kJ .
Gain of potential energy = mgh = 12 x 9.8 x 32 = 3.763 kJ
Decrease in kinetic energy KE = 1/2 x 12 ( 40² - 7² )
= 9.306 kJ
increase in internal energy ΔE = 70 kJ
ΔQ = ΔE + PE - KE + W , W is work done by the gas
Putting the values
24 = 70 + 3.763 - 9.306 + W
W = - 40.457 kJ .
Since W is negative that means work is done on the system .
types of aerobic activities?
Answer:
swimming, cycling, jump rope, brisk walking, gardening, jogging
Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.
Answer:
a. of liquid at the time bubbles first emerge slowly from the liquid.
Explanation:
Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.
How many planets on the solar system?
Answer:
8
Explanation:
tsijtsiztuztuistizrizturzurz
Answer:
8
Explanation:
Mercury, Venus, earth , Mars, jupiter, saturn , Uranus,Neptune
PLEASE HELP QUICK which statement describes a primary difference between an electromagnetic wave and mechanical wave?
A. electromagnetic waves can travel through empty space
B. electromagnetic waves can be transverse longitudinal or surface waves
C. electromagnetic waves can only travel through solids liquids or gases
D. electromagnetic waves need a medium to transfer energy
Answer:
A.
Explanation:
An electromagnetic wave is produced by the interaction between a variable electric field, and a magnetic electric field, which propagates in space, even in vaccuum, at a fixed speed, whilst the mechanical waves require a medium in order to transfer energy.Answer: A
Explanation:
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium withthe mixture. The piston is then moved inward very slowly, that thegas is always in thermal equilibrium with the ice-water mixture,what happens to the following(increase, decrease, same)?
a. volume of gas
b. temperature of gas
c. internal energy of gas,
d. pressure of gas
Answer:
a. volume of gas: (decreases)
b. temperature of gas: (same)
c. internal energy of gas: (same)
d. pressure of gas: (increases)
Explanation:
We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.
Then we put in a reservoir at 0°C (the mixture of water and ice)
remember that the state equation for an ideal gas is:
P*V = n*R*T
and:
U = c*n*R*T
where:
P = pressure
V = volume
n = number of mols
R = constant
c = constant
T = temperature.
Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.
Then in the equation:
P*V = n*R*T
all the terms in the left side are constants.
P*V = constant
And knowing that:
U = c*n*R*T
then:
n*R*T = U/c
We can replace it in the other equation to get:
P*V = U/c = constant.
Now, the piston is (slowly) moving inwards, then:
a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.
b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.
c) Internal energy of the gas:
we have:
P*V = n*R*T = constant
and:
P*V = U/c = constant.
Then:
U = c*Constant
This means that the internal energy does not change.
d) Pressure of the gas:
Here we can use the relation:
P*V = constant
then:
P = (constant)/V
Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.
And the quotient is equal to P.
Then if the volume decreases, we will see that the pressure increases.
A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2
Answer:
i think 692m/s2 is the correct answer
Why are carbon atoms able to form many organic compounds?
A. Carbon atoms have strong attraction to other elements.
B. Carbon atoms attract electrons from other atoms.
C. Carbon atoms can form many types of bonds with other carbon.
D. All of the above
Answer:
yo imma so I dunno find out yourself
Explanation:
dhdhdhdnndsisijjsksskskekekekkekssisisieieieiiwiwiwieiwieidjdjddi?
Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?
An electrically charged object
An uncharged object
A positively charged object
A negatively charged object
Answer:
its An uncharged object.
if its not charged the electrically wont go on it
Answer:
uncharged object
Explanation:
A motorcyclist is making an electric vest that, when connected to the motorcycle's 12 V battery, will warm her on cold rides. She is using 0.25-mm-diameter copper wire, and she wants a current of 4.2 A in the wire. Part A What length wire must she use
Answer:
L = 8.35 m
Explanation:
The lenght of a wire L can be calculated using the following expression:
L = R A/ρ (1)
Where:
R: resistance of the wire
A: Cross section area of the wire
ρ: resistivity of the copper wire.
With this expression we realize that we do not have the area of the cross section, and the resistance of the wire either.
To calculate the area we can use the following expression:
A = πr² (2)
If the diameter is 0.25 mm, then the radius is half, 0.125 mm. Converting this in meter it will have to be:
0.125 /1000 = 0.000125 m
Replacing we have:
A = π(0.000125)²
A = 4.91x10⁻⁸ m²
The reported resistivity of a copper wire is 1.68x10⁻⁸ Ω.m, so we just need to determine the resistance, which can be found using Ohm's law:
R = V/I (3)
Replacing (3) into (1) we have:
L = (V * A) / (I * ρ) (4)
So finally, the length of the copper wire will be:
L = (12 * 4.91x10⁻⁸) / (4.2 * 1.68x10⁻⁸)
L = 8.35 mHope this helps
who has brown hair and brown eyes but is a boy
Answer:
I have strawberry blonde/brown hair blue eyes and a girl lol
Explanation:
A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil can be poured into the glass to keep it still sinking? The density of the oil is 900 kg / m
Answer:
Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...
A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)
Answer:
h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0
The correct result is that of a positive height
Explanation:
For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man
v_s = y / t
t = [tex]\frac{y}{ v_s}[/tex]
this height is y = h
t = \frac{h}{ v_s}
the man has a response time of t = t₀, therefore
time to move is
t' = t - t₀
the initial height of flower pot is
y = y₀ + v₀ t' - ½ g t'²
when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0
0 = y₀ +0 - ½ g (t -t₀)²
if the initial height is i = h,
h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2
[tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²
[tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0
[tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]
To know the height, you must solve the second degree equation, it is much easier with numerical values.
The correct result is that of a positive height
Name the state of matter that diffusion happens the fastest in.
Answer:
Liquids
Explanation:
Diffusion occurs fastest in liquids.
A toy car can go 5 mph. How long would it take to go 12 miles?
PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.
The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.
Which number label represents the cell membrane?
1
2
4
6
(this is middle school science)
Answer:
1. cell membrane
2. golgi body
3. mitochondrion
4. cytoplasm
5. nucleolus
6. nucleus
Explanation:
The correct answer to this question is Option A; 6.
Why?
In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.
~Thank you~
HELPPPPP
What can you infer about the strength and direction of forces experienced by the pod and space station when they collided? What evidence from today’s activities supports your inference?
Answer:
In the collision, the strength of the force exerted on the pod is greater than the strength of the force exerted on the space station, but those forces are exerted in opposite directions.
Explanation:
Which statement best compares coal and ores?
Both are burned for energy.
Both take millions of years to form.
Both require oxygen to form.
Both are used to make coins.
Answer:
Option 2 both take millions of years to form
Explanation:
Both coal and ores take millions of years to form.
What are ores?Ore is a naturally occurring rock or silt that has precious minerals in it that may be extracted, processed, and sold for a profit. These minerals are usually metals. Mining is the process of removing ore from the soil. The valuable metals or minerals are then removed by treating or refining the ore, frequently through smelting.
The concentration of the desired ingredient in an ore is referred to as its grade. To decide if a rock has a high enough grade to be worth mining and is therefore regarded as an ore, the value of the metals or minerals it contains must be evaluated against the expense of extraction.
Typically, oxides, sulphides, silicates, or native metals like copper or gold are the minerals of interest. To separate the valuable components from the waste rock, ore must be treated. Numerous geological processes collectively known as ore genesis are responsible for the formation of ore deposits.
Learn more about ore here:
https://brainly.com/question/4712893
#SPJ2
3.
What part of your eye is responsible for regulating the amount of light that enters your eye?
Answer:
Iris
Explanation:
The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.
Answer:
I hope this helps.
Explanation:
A block of wood 3 cm on each
side has a mass of 27 g. What is the
density of the block? (Hint, don't
forget to find the volume of the
wood first using lx W h.)
Answer:
1g/cm3
Explanation:
volume of block is 3 cubed which is 27 cm3
we know density is m/v so d= 27g/27cm3
which is 1g/cm3
if my answer helps please mark as brainliest
In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position
Answer:
Explanation:
The relation between electric field and potential difference is as follows
E = - dV / dr
That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .
Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means
dV / dr = constant
E = constant
Electric field is constant .
So the option which is correct is
" points east and does not vary with position " .
Four electrons and one proton are at rest, all at an approximate infiitne distance away from each other. This original arrangment of the four particles is defined as having zero electrical potential energy No work is required to bring one electron from infitinty to a location defined as the origin, while the other three particles remain at infiniuty. This is because no voltage exists near the origin until the first electron arrives. (a) Now, with the first electron remaining fixed at the origin, how much work is required to bring one of the remaining electrons from infinity to the coordinate (0 m, 2.00 m)? The other three particles remain at infinity. If this second electron was subsequently released, how fast would it be traveling once it returned to infinity? (b) Nļw, considering the two electrons fixed 2.00 m apart, how much work is required to bring the third electron from infinity to the coordinate (3.00 m, 0 m)? The other two particles remain at infinity. If this third electron was subsequently released, how fast would it be traveling once it returned to infinity? (c) Now considering the three fixed electrons at the coordinates described above. How much work is required to bring the last electron from infinity to the coordinate (3.00 m, 4.00 m)? If this forth electron was subsequently released, how fast would it be traveling once it returned to infinity? (d) Now considering the three fixed electrons at the coordinates described above. Finally, how much work is required to bring the proton from infinity to a coordinate of (1.00 m, 1.00 m)? If the proton is subsequently released and we assume that minimum separation distance between a proton and an electron is 1.00 pm, then how fast will the proton be traveling once it crashes into an electron?
Answer:
a) W = 1.63 10⁻²⁸ J, b) W = 1.407 10⁻²⁷ J, c) W = 1.68 10⁻²⁸ J,
d) W = - 4.93 10⁻²⁸ J
Explanation:
a) In this problem we have an electron at the origin, work is requested to carry another electron from infinity to the point x₂ = 0, y₂ = 2.00m
If we use the law of conservation of energy, work is the change in energy of the system
W = ΔU = U_∞ -U
the potential energy for point charges is
U =k [tex]\sum \frac{q_i q_j}{r_{ij} }[/tex]
in this case we only have two particles
U = k [tex]\frac{q_1q_2}{r_{12} }[/tex]
the distance is
r₁₂ = [tex]\sqrt{(x_2-x_1)^2 + ( y_2-y_1)^2 }[/tex]
r₁₂ =[tex]\sqrt{ 0 + ( 2-0)^2}[/tex]Ra 0 + (2-0)
r₁₂ = √2= 1.4142 m
we substitute
W = k \sum \frac{q_i q_j}{r_{ij} }
let's calculate
W = [tex]\frac{ 9 \ 10^9 (1.6 \ 10^{-19})^2 }{1.4142}[/tex] 9 109 1.6 10-19 1.6 10-19 / 1.4142
W = 1.63 10⁻²⁸ J
b) the two electrons are fixed, what is the work to bring another electron to x₃ = 3.00 m y₃ = 0
in this case we have two fixed electrons
U = k [tex]( \frac{q_1q_3}{r_{13} } + \frac{q_2q_3}{r_{23} } )[/tex]
in this case all charges are electrons
q₁ = q₂ = q₃ = q
W = U = k q² [tex]( \frac{1}{r_{13} } + \frac{1}{r_{23} } )[/tex]
the distances are
r₁₃ = [tex]\sqrt{(3-0)^2 + 0}[/tex]RA (3.00 -0) 2 + 0
r₁₃ = 3
r₂₃ = [tex]\sqrt{ 3^2 + 2^2}[/tex]Ra (3 0) 2 + (2 0) 2
r₂₃ = √13
r₂₃ = 3.606 m
let's look for the job
W = U
let's calculate
W =[tex]{9 \ 10^3 ( 1.6 10^{-19})^2 }({\frac{1}{3} + \frac{1}{3.606} } )[/tex]
W = 1.407 10⁻²⁷ J
c) the three electrons are fixed, we bring the four electron to x₄ = 3.00m,
y₄ = 4.00 m
W = U = k [tex]( \frac{q_1q_4}{r_{14 }} + \frac{q_2q_4}{r_{24} } + \frac{q_3q_4}{r_{34} } )[/tex]
all charges are equal q₁ = q₂ = q₃ = q₄ = q
W = k q² [tex](\frac{1}{r_{14} } + \frac{1}{r_{24} } + \frac{1}{r_{34} } )[/tex]
let's look for the distances
r₁₄ = [tex]\sqrt{3^2 +4^2}[/tex]
r₁₄ = 5 m
r₂₄ = [tex]\sqrt{3^2 + ( 4-2)^2}[/tex]
r₂₄ = √13 = 3.606 m
r₃₄ = [tex]\sqrt{(3-3)^2 + (4-0)^2}[/tex]
r₃₄ = 4 m
we calculate
W = 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{5} + \frac{1}{3.606} + \frac{1}{4} )[/tex]
W = 1.68 10⁻²⁸ J
d) we take the proton to the location x5 = 1m y5 = 1m
W = U = k [tex]( \frac{q_1q_5}{r_{15} } + \frac{q_2q_5}{r_{25} } + \frac{q_3q_5}{r_{35} } + \frac{q_4q_5}{r_{45} } )[/tex]
in this case the charges have the same values but charge 5 is positive and the others negative, so the products of the charges give a negative value
W = - k q² [tex]( \frac{1}{r_{15} } + \frac{1}{r_{25} } + \frac{1}{r_{35} } + \frac{1}{r_{45} } )[/tex]
we look for distances
r₁₅ = [tex]\sqrt{ 1^2 +1^2}[/tex]Ra (1-0) 2 + (1-0) 2
r₁₅ = √ 2 = 1.4142 m
r₂₅ = [tex]\sqrt{ (2-1)^2 +1^2}[/tex]
r₂₅ = √2 = 1.4142 m
r₃₅ = [tex]\sqrt{ ( 3-1)^2 +1^2}[/tex]
r₃₅ = √5 = 2.236 m
r₄₅ = [tex]\sqrt{ (3-1)^2 + (4-1)^2}[/tex]
r₄₅ = √13 = 3.606 m
we calculate
W = - 9 10⁹ (1.6 10⁻¹⁹)² [tex]( \frac{1}{1.4142} +\frac{1}{1.4142} + \frac{1}{2.236} + \frac{1}{3.606} )[/tex]
W = - 4.93 10⁻²⁸ J
Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.
Answer:
The electric potential at the center of the meter stick is 54 KV.
Explanation:
Electric potential (V) is given as:
i.e V = [tex]\frac{kq}{r}[/tex]
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m
So that,
V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]
= [tex]\frac{2.7*10^{4} }{0.5}[/tex]
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these
Answer:
a) E = 0
b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
From which we have;
[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]
E = 0/A = 0
E = 0
b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]
[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]
By Gauss theorem, we have;
[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]
Therefore, we get;
[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]
The electrical field outside the spherical shell
[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]
[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]
Therefore, we have;
[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
Greatest to least order
Answer:
Explanation:
FBEDAC
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:
