Answer and Explanation:
The computation of the object distance related to two quantities is shown below:
It could find out by using the lens formula which is shown below:
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
where,
v = image distance
u = object distance
f = focal length
It could be found by applying the above formula i.e considering the image distance, object distance and the focal length
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.
Answer:
they must be affordable because they have to pay for it or they wont get the stuff they are bying.
Explanation:
need a brainliest please.
Answer: B, they must be affordable.
Explanation:
Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?
Answer:
θ = 45.15°
Explanation:
We need to use the grating equation in this question. The grating equation is given as follows:
mλ = d Sin θ
where,
m = order number = 3
λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m
d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m
θ = angle from the axis = ?
Therefore,
(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ
Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)
Sin θ = 0.709
θ = Sin⁻¹(0.709)
θ = 45.15°
Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?
Answer:a) 4+8+24=36
B) 1/4+1/8+1/24=10
C) yu will connect them in parallel connection.
D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.
Explanation:
(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.
(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
What is resistance?Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;
[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]
The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.
[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]
(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.
(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.
Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.
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Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
6. If you wanted to develop a telescope, what kind of lenses would you use for the objective lens (the lens that collects the light) and the eyepiece? Explain your reasoning. Draw a picture with ray tracing of your setup.
Answer:
objetive: a converging lens for large diameter lenses
eyepiece you must select a lens with a small focal length and the diameter is not important
The selected lenses should decrease chromatic aberration.
Explanation:
A telescope is an instrument that collects light from very distant objects, therefore very weak.
Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.
For the eyepiece you must select a lens with a small focal length and the diameter is not important
the telescope magnification is
m = f_objective / F_ocular
The selected lenses should decrease chromatic aberration.
In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Explanation:
Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition
Answer:
The maximum wavelength of the e-m wave is 6.9 x 10^-7 m
Explanation:
Energy required to trigger a response = 1.8 eV
we convert to energy in Joules.
1 eV = 1.602 x 10^-19 J
1.8 eV = [tex]x[/tex] J
[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J
The energy of an electromagnetic wave is gotten as
E = hf
where
h is the Planck's constant = 6.63 x 10^-34 J-s
and f is the frequency of the wave.
substituting values, we have
2.88 x 10^-19 = 6.63 x 10^-34 x f
f = (2.88 x 10^-19)/(6.63 x 10^-34)
f = 4.34 x 10^14 Hz
We know that the frequency of an e-m wave is given as
f = c/λ
where
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength of the e-m wave
From this we can say that
λ = c/f
λ = (3 x 10^8)/(4.34 x 10^14)
λ = 6.9 x 10^-7 m
The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3
Answer:
1) M = 11.3V2) 195.49 gramsExplanation:
1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;
M ∝ V
M = kV
For every proportionality sign, there will always be a proportionality constant 'k'
Since the proportionality constant is the density (D) of the metal, the equation will become;
M = DV
Given the density to be 11.3 g/cm3, the equation will become;
M = 11.3V
Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V
2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;
M = 11.3V
M = 11.3 * 17.3
M = 195.49 grams
Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
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550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.
Required:
a. What is its frequency?
b. What type of electromagnetic radiation might this be?
Answer:
a) 5.5×10^17 Hz
b) visible light
Explanation:
Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;
λ= 5.5 × 10^-10 m
Since;
c= λ f and c= 3×10^8 ms-1
f= c/λ
f= 3×10^8/5.5 × 10^-10
f= 5.5×10^17 Hz
The electromagnetic wave is visible light
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low- mass sail and the energy and momentum of sunlight for propulsion.
Should the sail be absorbing or reflective? Why?
a. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is smaller than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
b. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is larger than for reflective sail, therefore the radiation pressure is larger for the absorbing sail
c. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is smaller than for reflective sail, therefore the radiation pressure is larger for the absorbing sail.
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
Answer:
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?
Answer:
Explanation:
Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/
α = 26.56°
Laser angle with Normal = 90 - 26.56 = 63.44 °
Assuming a red laser, refractive index in water is 1.331.
Angle of refraction in water is given by:
Ref Ind = Sin i / Sin r
1.331 = Sin 63.44 / Sin r
Sin r = 0.8945 / 1.331 = 0.6721
Angle r = 42.22°
For the path in water:
Tan 42.22 = x / 3.2
x = 2.9m where x is the lateral displacement of the laser ince it hits the water
So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool
6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion
Answer:
C. planetary accretion
Explanation:
Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.
Answer:
[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]
Explanation:
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.
Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.
A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N
Answer:
The correct answer is 500 N
Explanation:
This is an exercise in Newton's third law or law of action and reaction
The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.
Which force we call action does not matter, the analysis and conclusions are the same
The correct answer is 500N
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Answer:
b) True. potencial diferencie does not depend on orientation
Explanation:
In this exercise we are asked to show which statements are true.
The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.
It does not refer to the height of the system.
We can now review the claims
a) False. Potential not to be refers to height
b) True. Does not depend on orientation
c) False The potential does not refer to the altitude but to the Earth's charge
Terms to describe the opposition by a material.to being magnetised is
Answer:
Repulsion
Explanation:
How could a country benefit from making it into space?
Answer:
space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
