Answer:
The answer is option b.the buoyant force is greater on the rock in water.
If you were to calculate the pull of the Sun on the Earth and the pull of the Moon on the Earth, you would undoubtedly find that the Sun's pull is much stronger than that of the Moon, yet the Moon's pull is the primary cause of tides on the Earth. Tides exist because of the difference in the gravitational pull of a body (Sun or Moon) on opposite sides of the Earth. Even though the Sun's pull is stronger, the difference between the pull on the near and far sides is greater for the Moon.
Required:
a. "Let F(r) be the gravitational force exerted on one mass by a second mass a distance r away. Calculate dF(r)/dr to show how F changes as r is changed.
b. Evaluate this expression for dF(r) jdr for the force of the Sun at the Earth's center and for the Moon at the Earth's center.
c. Suppose the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun. Is there any point where dF(r)/dr for the two forces has the same value?
Answer:effective
Explanation:
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.01 kg of this oil from 23 °C to 191 °C?
Answer:
Q = 590,940 J
Explanation:
Given:
Specific heat (c) = 1.75 J/(g⋅°C)
Mass(m) = 2.01 kg = 2,010
Change in temperature (ΔT) = 191 - 23 = 168°C
Find:
Heat required (Q)
Computation:
Q = mcΔT
Q = (2,010)(1.75)(168)
Q = 590,940 J
Q = 590.94 kJ
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is
Answer:
the frequency of revolution of the second particle is f
Explanation:
centripetal force is balanced by the magnetic force for object under magnetic field is given as
Mv²/r= qvB
But v= omega x r
Omega= 2pi x f
f= qB/2pi x M
So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is:
Answer:
Three halves of a wavelength I.e 7lambda/2
Explanation:
See attached file pls
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/schange in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius
Answer:
4519.60 J/KExplanation:
Change in entropy is expressed as ΔS = ΔQ/T where;
ΔQ is the total heat change during conversion of ice to water.
T is the room temperature
First we need to calculate the total change in heat using the conversion formulae;
ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)
m is the mass of ice/water = 900g = 0.9kg
L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg
c is the specific heat capacity of water = 4200J/kgK
Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K
Substituting the given values into ΔQ;
ΔQ = 0.9(333000)+0.9(4200)(283)
ΔQ = 299700 + 1069740
ΔQ = 1,369,440 Joules
Since Change in entropy ΔS = ΔQ/T
ΔS = 1,369,440/30+273
ΔS = 1,369,440/303
ΔS = 4519.60 J/K
Hence, the change in entropy of the universe is 4519.60 J/K
The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of the light passing through it. At wavelength 486.1 nm (blue, designated with letter F) -> nF=1.497, and at wavelength 656.3 nm (red, designated with letter C) -> nC=1.488. Two beams (one of each wavelength) are prepared to coincide, and enter the flat polished surface of an acrylic block at angle of 45 arc degree measured from the normal to the surface. What is the angle between the blue beam and the red beam in the acrylic block?
Answer:
The angle between the blue beam and the red beam in the acrylic block is
[tex]\theta _d =0.19 ^o[/tex]
Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is [tex]n_F = 1.497[/tex]
The wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]
The refractive index of the transparent acrylic plastic for red light is [tex]n_C = 1.488[/tex]
The wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]
The incidence angle is [tex]i = 45^o[/tex]
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
[tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]
[tex]r_F = 28.18^o[/tex]
Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
[tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]
[tex]r_F = 28.37^o[/tex]
The angle between the blue beam and the red beam in the acrylic block
[tex]\theta _d = r_C - r_F[/tex]
substituting values
[tex]\theta _d = 28.37 - 28.18[/tex]
[tex]\theta _d =0.19 ^o[/tex]
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.
Answer:
The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 11^o[/tex]
The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as
[tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]
Where [tex]\beta[/tex] is mathematically represented as
[tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]
[tex]\beta = 708.1 \ rad[/tex]
So
[tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
3. Identify the mathematical relationship that exists between pressure and volume, when temperature and quantity are held constant, as being directly proportional or inversely proportional. Explain your answer and write an equation that relates pressure and volume to a constant, using variables
Answer:
P = cte / V
therefore pressure and volume are inversely proportional
Explanation:
For this exercise we can join the ideal gases equation
PV = n R T
they indicate that the amount of matter and the temperature are constant, therefore
PV = cte
P = cte / V
therefore pressure and volume are inversely proportional
A car is moving along a road at 28.0 m/s with an engine that exerts a force of
2,300.0 N on the car to balance the drag and friction so that the car maintains a
constant speed. What is the power output of the engine?
Answer:
Power = Force × Distance/time
Power = Force × Velocity
Power = 2,300.0 N × 28.0 m/s²
Power = 64400 Nm/s
Explanation:
First show the formula of Power
Re-arrange formula and used to work out Power
Pretty simple stuff!
Hope this Helps!!
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).
Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
Explanation:
Thermal kinetic energy of electron or proton = KE
∴ KE = 3kbT/2
given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K
so we substitute
KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2
kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )
Now we know that
mass of electron M'e = 9.109 × 10⁻³¹
mass of proton M'p = 1.6726 × 10⁻²⁷
We also know that
KE = p₂ / 2m
from the equation, p = √ (2mKE)
{ p is momentum, m is mass }
de Broglie wavelength = β
so β = h / p = h / √ (2mKE)
h = Planck's constant = 6.626 × 10⁻³⁴
∴ βe = h / √ (2m'e × KE)
βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )
βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰
βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵
βe = 2.443422 × 10⁻⁹ m
βp = h / √ (2m'p ×KE)
βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )
βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶
βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³
βp = 5.702140 × 10⁻¹¹ m
When static equilibrium is established for a charged conductor, the electric field just inside the surface of the conductor is
Answer:
The electric field just inside the charged conductor is zero.
Explanation:
Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.
For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.
WILL MARK BRAINLIEST!!An igneous rock has large red, black, and green crystals. How else can this rock be accurately described?
O fine texture
O cooled quickly
O intrusive origin
O created by lava
Answer:
D
Explanation:
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?
Answer:
4.31 N
Explanation:
Given:
Δy = -55 m
v₀ = 0 m/s
v = -29 m/s
Find: a
v² = v₀² + 2aΔy
(-29 m/s)² = (0 m/s)² + 2a (-55 m)
a = -7.65 m/s²
Sum of forces in the y direction:
∑F = ma
R − mg = ma
R = m (g + a)
R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)
R = 4.31 N
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.
Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still connected to the battery. After this is done, we find that
Answer:
The voltage across the capacitor will remain constant
The capacitance of the capacitor will increase
The electric field between the plates will remain constant
The charge on the plates will increase
The energy stored in the capacitor will increase
Explanation:
First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.
The charge on the capacitor is equal to
Q = CV
Since the voltage is constant, and the charge increases, the capacitance will also increase.
The energy in a capacitor is given as
E = [tex]\frac{1}{2}CV^{2}[/tex]
since the capacitance has increased, the energy stored will also increase.