Answer:
t = 5.03 s
Explanation:
To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.
You use the following formula:
[tex]\omega = \frac{v}{r}[/tex]
w: angular speed
v: tangential speed
r: radius of the trajectory = 8 m
For you have:
[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]
For her brother:
[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]
Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:
[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]
They encounter for θ = 2π-θ':
[tex]\omega t=2\pi-\omega' t[/tex]
You replace the values of the parameters in the previous equation and solve for t:
[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]
Hence, Sheila encounter her brother in 5.03 s
Which of the following is analogous to the pipes in an electrical circuit?
A. capacitors storing the incoming charge from the battery
B. large resistors causing restrictions to the flow of charge
C. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery electric current being forced uphill by the battery
D. electric current being forced uphill by the battery back to the positive terminal
The correct answer is D. electric current being forced uphill by the battery back to the positive terminal.
What is Electric Current?
Electric current is the flow of electric charge through a conducting medium, such as a wire, due to the movement of electrons or ions. The flow of charge is typically caused by the presence of an electric field that creates a potential difference (voltage) between two points in a circuit
In an electrical circuit, pipes are analogous to wires or conductive paths that allow the flow of electric current. The flow of electric current is from the positive terminal of the battery to the negative terminal, which is opposite to the direction of conventional current flow. Therefore, option C is incorrect.
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The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?
Answer:
17.85°
Explanation:
To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
n1: index of refraction of Sophia's cornea = 1.387
n2: index of refraction of aqueous fluid = 1.36
θ1: angle to normal in the first medium = 17.5°
θ2: angle to normal in the second medium
You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:
[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]
hence, the angle to normal in the aqueous medium is 17.85°
Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a price ceiling for a bottle of water equal to the market equilibrium price during normal times. After all, it seems unfair that sellers of water gain because of a hurricane.
Answer:idk
Explanation:idk
Answer:
shortage of 50 water bottles
$2
30
Explanation:
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.653 kJ of heat. It shrinks on cooling, and the atmosphere does 389 J of work on the balloon. Express your answer to three significant figures and include the appropriate units.
Answer:
ΔE = ‒0.271 kJ
Explanation:
Let's begin by listing out the given variables:
q = -0.653 kJ, w = 0.389 kJ
Using the formula ΔE = q + w
ΔE = -0.653 + 0.388
ΔE = (‒0.655 + 0.382) kJ
ΔE = ‒0.271 kJ
Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?
Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]
The rms voltage (in V) measured across the secondary coil is calculated as;
[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Answer:
a.[tex]5.66ms^{-2}[/tex]
b.55 m
Explanation:
We are given that
Mass ,m=0.9 kg
Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]
1 m=100 cm
[tex]\theta=30^{\circ}[/tex]
R=55 m
a.Centripetal acceleration
[tex]a_c=gtan\theta[/tex]
[tex]a_c=9.8tan30^{\circ}[/tex]
[tex]a_c=5.66 m/s^2[/tex]
Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]
b. Radius of curve,R=55 m
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg
Answer:
(possibly) Box D
Explanation:
The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.
Which formation is one feature of karst topography?
Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.
What is karst topography?Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.
The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.
When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.
Hence, sinkholes formation is one feature of karst topography
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Answer: A) Caves
Explanation:
4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Answer:
Explanation:
The work required to push will be equal to work done by friction . Let d be the displacement required .
force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction
work done by force of friction
mg x μ x d = 660
80 x 9.8 x .272 x d = 660
d = 3 .1 m .
70 pointss yall !!! helpp
Answer:
A= The type of plant
B= How tall the plant is
Explanation:
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
Answer:-
-1 m/s^2
Explanation:
The average acceleration is given by dividing the change in velocity by change in time;
[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]
[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]
the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.
1. An object with a mass of 15 kilograms is pushed by a force of 30 Newtons. How much does
it accelerate?
Answer: [tex]2m/s^2[/tex]
Explanation:
[tex]Formula: F=ma[/tex]
Where;
F = force
m = mass
a = acceleration
Solve for a;
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?
Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s
Explanation: Please see the attachment below
The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
The given parameters:
Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 radThe angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]
where;
[tex]\omega_i[/tex] is the initial angular velocity
[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]
Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
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Which of the following best describes the current age of the Sun?
A.) It is near the end of its lifespan.
B.) It is about halfway through its lifespan.
C.) It is early in its lifespan.
D.) We do not have a good understanding of the Sun's age.
Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?
Answer:
Ff = 33.4N
Explanation:
To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.
The frictional force is given by:
[tex]F_f=\mu_kN[/tex] (1)
Ff: frictional force = ?
µk: coefficient of kinetic friction = 0.167
N: normal force of the object = 200N
You replace the values of the parameters in the equation (1):
[tex]F_f=(0.167)(200N)=33.4N[/tex]
The frictional force, while the objects is moving, is 33.4N
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?
Explanation:
The relation between resistance and resistivity is given by :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]\rho[/tex] is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to
Answer:
The time when the ball strikes the ground is closest to [tex]t_t = 9.4 \ s[/tex]
Explanation:
From the question we are told that
The time of projection is t = 0.0 s
The distance of the point from the ground is [tex]d = 90 \ m[/tex]
The initial velocity of the ball is [tex]v _i = 36 .2 \ m/s[/tex]
generally the time required to reach maximum height is
[tex]t_r = \frac{g}{v}[/tex]
Where is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
Substituting values
[tex]t_r = \frac{36.2}{9.8}[/tex]
[tex]t_r = 3.69 s[/tex]
when returning the time and velocity at the roof level is t = 3.69 s and u = 36.2 m/s this due to the fact that air resistance is negligible
The final velocity at which it hit the ground is
[tex]v_f^2 = u^2 + 2ag[/tex]
So
[tex]v_f = \sqrt{ u^2 + 2gs}[/tex]
substituting values
[tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]
[tex]v_f = 55.45 \ m/s[/tex]
The time taken for the ball to move from the roof level to the ground is
[tex]t_g = \frac{v-u}{a}[/tex]
substituting values
[tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]
[tex]t_g = 1.96 \ s[/tex]
The total time for this travel is
[tex]t_t = t_g + 2 t_r[/tex]
[tex]t_t = 1.96 + 2(3.69)[/tex]
[tex]t_t = 9.4 \ s[/tex]
Of one of the planets becomes a black hole , what would the escape speed be?
Answer:
If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.
Explanation:
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Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
Answer:
[tex]\large \boxed{42\, \mu \text{C}}$[/tex]
Explanation:
The formula for the force exerted between two charges is
[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]
where k is the Coulomb constant.
The charges are identical, so we can write the formula as
[tex]F=k\dfrac{q^{2}}{r^2}[/tex]
[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]
Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
