A theory does not change into a scientific law with the accumulation of new or better evidence. A theory will always remain a theory; a law will always remain a law. Both theories and laws could potentially be falsified by countervailing evidence. Theories and laws are also distinct from hypotheses.
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.
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to easily count them. To spread out the fringe pattern, one could
Answer:
halve the slit separation
Explanation:
As we know that
In YDS experiment, the equation of fringe width is as follows
[tex]\beta = \frac{\lambda D}{d}[/tex]
where,
D denotes the separation in the middle of screen and slits
d denotes the distance in the middle of two slits
And to increase the Δx we have to decrease the d i.e, the distance between the two slits
Hence, the first option is correct
A fish is 11.9 cm from the front surface of a fish bowl of radius 33 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl
Answer:
The fish would appear 42.7 cm on the left side from the front of the bowl.
Explanation:
The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.
But, f = [tex]\frac{radius of curvature}{2}[/tex]
= [tex]\frac{33}{2}[/tex]
f = 16.5 cm
Substitute f = 16.5 = [tex]\frac{165}{10}[/tex], and u = 11.9 = [tex]\frac{119}{10}[/tex] in equation 1;
[tex]\frac{10}{165}[/tex] = [tex]\frac{10}{119}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{10}{165}[/tex] - [tex]\frac{10}{119}[/tex]
= [tex]\frac{1190 - 1650}{19635}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{-460}{19635}[/tex]
⇒ v = [tex]\frac{19635}{-460}[/tex]
= -42.6848
v = 42.7 cm
The fish would appear 42.7 cm on the left side from the front of the bowl.
The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. experiment hypothesis observation control
The correct answer is C. Observation
Explanation:
An observation is a statement a describes a phenomenon, which is the result of measuring the phenomenon or using the senses to collect information about it. Additionally, observations are part of the Scientific method because through observations it is possible to understand phenomena.
The sentence presented is an observation because this statement is the result of the researcher observing or measuring how fast kernels pops, which means the statement derives from studying a phenomenon. Also, this cannot be classified as a hypothesis because a hypothesis is a probable explanation, and it cannot be classified as an experiment because the experiment is the general method to prove or disprove a hypothesis.
Metal 1 has a larger work function than metal 2. Both are illuminated with the same short-wavelength ultraviolet light.
Do electrons from metal 1 have a higher speed, a lower speed, or the same speed as electrons from metal 2? Explain.
Answer:
a lower speed
Explanation:
Let us look closely at the Einstein's photoelectric equation;
KE= E-Wo
Where;
KE= kinetic energy of the emitted photoelectron
E= energy of the incident photon
Wo= work function of the metal
Hence,where Wo for metal 1 > Wo for metal 2, it follows that KE for metal 1 must also be less than KE for metal 2.
This is because the difference between E and Wo for metal 1 is smaller than the same difference for metal 2 hence the answer.
A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
A string of holiday lights has 15 bulbs with equal resistances. If one of the bulbs
is removed, the other bulbs still glow. But when the entire string of bulbs is
connected to a 120-V outlet, the current through the bulbs is 5.0 A. What is the
resistance of each bulb?
Answer:
Resistance of each bulb = 360 ohms
Explanation:
Let each bulb have a resistance r .
Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.
[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]
[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]
R(equivalent) = r/15
Now, As per Ohms Law :
V = I * R(equivalent)
120 V = 5 A * r/15
r = 360 ohms
A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.
Answer:
24.34 m/s
Explanation:
recall that one of the equations of motions takes the form:
v = u + at
where,
v = final velocity
u = initial velocity (given as 22.2 m/s)
a = acceleration (given as 2.68m/s²)
t = time elapsed during acceleration (given as 0.80s)
since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:
v = u + at
v = 22.2 + (2.68) (0.8)
v = 24.34 m/s
A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.
Answer:
The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Explanation:
Given that,
A beam of light from a laser illuminates a glass.
Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]
Index of refraction is 2.83.
We need to calculate the speed of light pulse in glass
Using formula of speed
[tex]v=\dfrac{c}{\mu}[/tex]
Put the value into the formula
[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]
[tex]v=1.06\times10^{8}\ m/s[/tex]
We need to calculate the time of short pulse of light beam
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]
[tex]t=2.37\times10^{-9}\ sec[/tex]
Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?
Assuming constant speeds, the P-wave covers a distance d in time t such that
9000 m/s = d/(60 t)
while the S-wave covers the same distance after 1 more minute so that
5000 m/s = d/(60(t + 1))
Now,
d = 540,000 t
d = 300,000(t + 1) = 300,000 t + 300,000
Solve for t in the first equation and substitute it into the second equation, then solve for d :
t = d/540,000
d = 300,000/540,000 d + 300,000
4/9 d = 300,000
d = 675,000
So the earthquake center is 675,000 m away from the seismic station.
A converging lens 7.50 cm in diameter has a focal length of 330 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of resolving power of the human eye. Part A If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.10 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 600 nm
Answer:
D Is 430m
Explanation:
See attached file
Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?
Answer:
The energy contained in a single pulse is 90,200 J.
Explanation:
Given;
power of the laser, P = 82 TW = 82 x 10¹² W
time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s
the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m
The energy contained in a single pulse is calculated as;
E = Pt
where;
P is the power of each laser
t is the time to generate the power
E = (82 x 10¹²)(1.1 x 10⁻⁹)
E = 90,200 J
Therefore, the energy contained in a single pulse is 90,200 J
. Two charges, Q1 and Q2 , are separated by a certain distance R. If the magnitudes of the charges are halved, and their separation is also halved, then what happens to the electrical force between these charges
Answer: Magnitude of electrical force stays the same.
Explanation:
Equation:
[tex]F_{e} =k\frac{Q_{1}Qx_{2} }{r^{2} }[/tex]
Since the magnitude of each charge is halved.
and
the separation is halved.
[tex]F_{e} =k\frac{(.5Q_{1}*.5Q_{2} }{(.5r)^{2} }[/tex]
[tex]F_{e} =k\frac{.25*Q_{1}Qx_{2} }{.25*r^{2} }[/tex]
Cancel out .25 on the numerator and denominator. Leaving the original equation.
Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).
Answer:
The water pressure on the upper pipe is 92.5 kPa.
Explanation:
Given that,
Pressure in lower pipe= 120 kPa
Speed of water in lower pipe= 1 m/s
Acceleration due to gravity = 10 m/s²
Density of water = 1000 kg/m³
Radius of lower pipe = 12 m
Radius of uppes pipe = 6 m
Height of upper pipe = 2 m
We need to calculate the velocity in upper pipe
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]
[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]
[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]
[tex]v_{2}=4\ m/s[/tex]
We need to calculate the water pressure on the upper pipe
Using bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Put the value into the formula
[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]
[tex]120500=P_{2}+28000[/tex]
[tex]P_{2}=120500-28000[/tex]
[tex]P_{2}=92500\ Pa[/tex]
[tex]P_{2}=92.5\ kPa[/tex]
Hence, The water pressure on the upper pipe is 92.5 kPa.
what happens to the weight of the body when it is falling freely under the action of gravity
Answer:
A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object
Explanation:
plz mark me as brainliest
Answer:
Gradually increases until the maximum weight reaches the surface of the earth
Explanation:
1 hallar el trabajo mecanico de un cuerpo que tiene una fuerza de 250 newton y recorre 750 metros
2 hallar la potencia necesaria para levantar un transformador de masa 2500kg,una altura de 4 metros en un tiempo de 30 segundos
porfa es para hoy
Answer: TRACK
Explanation:
When using science to investigate physical phenomena, which characteristic of the event must exist? predictable repeatable provable readable
Answer:
Not sure but I believe predictable.
Explanation:
Phenomena usually consists of :
- A history, a date in which the physical phenomenon has occurred.
- A source, a place or reason to why or where the physical phenomena has occured.
According to this, I want to say predictable.
It is not repeatable, there are one-time phenomenons that have occurred that scientists to this day still have not recorded again such as the Big Bang.
It is not provable. Most of the theories earlier scientists and historians have predicted have not yet been proved. Yet rather, somehow, they have been explored and investigated.
It is not readable. This is self explanatory, some things scientists investigate are not written down, nor read. It starts with a mental theory and then immediately goes to the next phase of investigation.
What's the minimum Out PUT WORK
required to raise 14,0m3 of water 26.0m?
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
y=k/x, x is halved.
what happens to the value of y
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if everyone chips in for gas. Someone asks how much the gas will cost per person on a round trip. Your first step is to call your smarter brother to see if he'll figure it out for you. Naturally he's too busy to bother, but he does tell you that it is 2015 km to Mexico City, there's 11 cents to the peso, and gas costs 5.8 pesos per liter in Mexico. You know your car gets 21 miles to the gallon, but we still don't have a clue as to how much the trip is going to cost (in dollars) each person in gas ($/person).
Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Answer:
6000 counts per secondExplanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m
Answer:
Explanation:
volume of water being lifted
= π r² h , where r is radius of cylinder and h is height of cylinder
= 3.14 x5² x 10
= 785 m³
mass of water = 785 x 10³ kg
mass of this much of water is lifted so that its centre of mass is lifted by height
10 / 2 = 5m .
So work done = mgh , m is mass of water , h is displacement of centre of mass and g is acceleration due to gravity
= 785 x 10³ x 9.8 x 5
= 38.465 x 10⁶ J
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d 1.53 cm and a plate area of A = 25.0 cm2. when the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
(a) Calculate the charge on the plates in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before Qi = _____
after Qf = ______
(b) Determine the capacitance (in F) and potential difference (in V) after immersion
(c) Determine the change in energy (in n]) of the capacitor Δυ = nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 275 V potential difference
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
Determine the capacitance (in F) and potential difference (in V) after immersion
Determine the change in energy (in nJ) of the capacitor AU nJ
Answer:
a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,
c) U = 54.7 nJ , d) ΔU = 54 nJ,
Explanation:
a) The capacity of a capacitor is defined
C = Q / V
Q = C V
can also be calculated using geometry consideration
C = e or A / d
we reduce to the SI system
A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²
d = 1.53 cm = 1.53 10⁻² m
we substitute
Q = eo A / d V
Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275
Q = 3.9757 10⁻¹⁰ C
let's reduce to pC
Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)
Q = 397.57 pC
when the capacitor is introduced into the water the dielectric constant is different
Q = k Q₀
Q = 80 397.57
Q = 3.18 104 pC
b) Find capacitance and voltage after submerged in water
C = k C₀
C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²
C = 1.157 10⁻¹⁰ F
V = Vo / k
V = 275/80
V = 3.4375 V
c) The stored energy is
U = ½ C V²
U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²
U = 5.47 10⁻⁸ J
let's reduce to nJ
109 nJ = 1 J
U = 54.7 nJ
d) energy after submerging
U = ½ (kCo) (Vo / k) 2
U = ½ Co Vo2 / k
U = U₀ / k
U = 54.7 / 80 nJ
U = 0.68375 nJ
the energy change is
ΔU = U₀ -U
ΔU = 54.7 - 0.687375
(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.
(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.
(c) The change in energy of the capacitor is 54.02 nJ.
Charge on the plate before immersionThe charge on the plate is calculated as follows;
[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]
Charge on the plate after immersion[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]
Capacitance and potential difference after immersion[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]
[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]
Change in energy of the capacitorThe initial energy of the capacitor is calculated as follows;
[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]
The final energy of the capacitor is calculated as follows;
[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]
Change in energy is calculated as follows;
[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]
Learn more about energy stored in a capacitor here: https://brainly.com/question/13578522
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)
Answer:
(a)106.4C
b)0.5676mm
Explanation:
(a)To get the charge that have passed through the starter then The current will be multiplied by the duration
I= current
t= time taken
Q= required charge
Q= I*t = 140*0.760 = 106.C
(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)
diameter of the conductor is 4.20 mm
But Radius= diameter/2= 4.20/2=
The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then
radius of the conductor is 0.0021m.
We can now calculate the area of the conductor which is
A = π*r^2
= π*(0.0021)^2 = 13.85*10^-6 m^2
We can proceed to calculate the current density below
J = 140/13.85*10^-6 = 10108303A/m
According to the listed reference:
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s
Therefore , the distance traveled is:
x = v*t = 0.7468 * 0.760 = 0.5676mm
(a) The charge passes through the starter motor is 106.4C.
(b) An electron travel along the wire while the starter motor is on 0.5676mm.
ElectronAnswer (a)
I= current
t= time taken
Q= required charge
Q= I*t
Q= 140*0.760
Q= 106.C
Answer (b)
The n electron travel along the wire while the starter motor is on:
Diameter of the conductor is 4.20 mm
Radius= diameter/2= 4.20/2
Radius =2.1mm
Radius of the conductor is 0.0021m.
A = π*r^2
A= π*(0.0021)^2
A= 13.85*10^-6 m^2
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )
Vd =0.0007468m/s
Vd =0 .7468 mm/s
The distance traveled is:
x = v*t
x= 0.7468 * 0.760
x = 0.5676mm
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What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?
Answer:
7.86×10⁻⁷ m
Explanation:
Using,
v = λf.................. Equation 1
Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.
make λ the subject of the equation
λ = v/f............... Equation 2
Note: All electromagnetic wave have the same speed which is 3×10⁸ m/s.
Given: f = 3.818×10¹⁴ Hz
Constant: v = 3×10⁸ m/s
Substitute these values into equation 2
λ = 3×10⁸/3.818×10¹⁴
λ = 7.86×10⁻⁷ m
Hence the wavelength of the electromagnetic radiation is 7.86×10⁻⁷ m
The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]
Given the following data:
Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]Scientific data:
Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]
To determine the wavelength of this electromagnetic radiation:
Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;
[tex]Wavelength = \frac{Speed }{frequency}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]
Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.
Answer:
[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]
Explanation:
The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:
[tex]\Delta \theta = \omega \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Steady angular speed, measured in radians per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:
[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]
[tex]\Delta \theta = 3116.11\,rad[/tex]
The change in angular displacement, measured in revolutions, is given by the following expression:
[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]
[tex]\Delta \theta = 495.944\,rev[/tex]
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
what is liquid pressure and its si unit?
The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.