Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices
Answer:
0.05cos10t
Explanation:
X(t) = Acos(wt+φ)
The oscillation angular frequency can be calculated using below formula
w = √(k/M)
Where K is the spring constant
But we were given body mass of 5.0 kg
We know acceleration due to gravity as 9.8m)s^2
The lenghth of spring which stretches =10 cm
Then we can calculate the value of K
k = (5.0kg*9.8 m/s^2)/0.10 m
K= 490 N/m
Then if we substitute these values into the formula above we have
w = √(k/M)
w = √(490/5)
= 9.90 rad/s=10rads/s(approximately)
Its position as a function of time can be calculated using the below expresion
X(t) = Acos(wt+φ)
We were given amplitude of 5 cm , if we convert to metre = 0.05m
w=10rads/s
Then if we substitute we have
X(t)=0.05cos(10×t)
X(t)= 0.05cos10t
Therefore,Its position as a function of time=
X(t)= 0.05cos10t
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
an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the device
Answer: R=24.2Ω
Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
[tex]P=\frac{V^{2}}{R}[/tex]
The resistance of the system is:
[tex]P=\frac{V^{2}}{R}[/tex]
[tex]R=\frac{V^{2}}{P}[/tex]
[tex]R=\frac{110^{2}}{500}[/tex]
R = 24.2Ω
For the device, resistance is 24.2Ω.
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 laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) about 20 times per second.part a. Determine the number of photons in one 80-mJ pulse.part b. Determine the average power of photons in one 80-mJ pulse during 1 s.
Answer:
a
[tex]n = 1.119 *10^{18} \ photons[/tex]
b
[tex]P = 1.6 \ W[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 2780 nm = 2780 *10^{-9} \ m[/tex]
The energy is [tex]E = 80 mJ = 80 *10^{-3} \ J[/tex]
This energy is mathematically represented as
[tex]E = \frac{n * h * c }{\lambda }[/tex]
Where c is the speed of light with a value [tex]c = 3.0 *10^{8} \ m/s[/tex]
h is the Planck's constant with the value [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
n is the number of pulses
So
[tex]n = \frac{E * \lambda }{h * c }[/tex]
substituting values
[tex]n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }[/tex]
[tex]n = 1.119 *10^{18} \ photons[/tex]
Given that the pulses where emitted 20 times in one second then the period of the pulse is
[tex]T = \frac{1}{20}[/tex]
[tex]T = 0.05 \ s[/tex]
Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as
[tex]P = \frac{E}{T}[/tex]
substituting values
[tex]P = \frac{ 80 *10^{-3}}{0.05}[/tex]
[tex]P = 1.6 \ W[/tex]
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]
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:
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Answer:
Gradually increases until the maximum weight reaches the surface of the earth
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.
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
write any two importance of gravitational force
Answer:
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Explanation:
The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.
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
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
Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over its entire length to a temperature of 600C. Calculate the heat transferred from the plate.
Answer:
Heat transferred = 22.9 watt
Explanation:
Given that:
[tex]T_1[/tex] = 27°C = (273 + 27) K = 300 K
[tex]T_2[/tex]= 600°C = (600 +273) K = 873 K
speed v = 2 m/s
length x = 40 cm = 0.4 cm
width = 1 cm = 0.001 m
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
From the tables of properties of air, the following values where obtained.
[tex]k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg[/tex]
To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:
reynolds number = [tex]\dfrac{\rho \times v \times x }{\mu}[/tex]
reynolds number = [tex]\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}[/tex]
reynolds number = [tex]\dfrac{0.98}{18.6 \times 10^{-6}}[/tex]
reynolds number = 52688.11204
Prandtl number = [tex]\dfrac{c_p \mu}{k}[/tex]
Prandtl number = [tex]\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}[/tex]
Prandtl number = [tex]\dfrac{0.018693}{0.02476}[/tex]
Prandtl number = 0.754963
The nusselt number for this turbulent flow over the flat plate can be computed as follows:
Nusselt no = [tex]\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}[/tex]
[tex]\dfrac{h \times 0.4}{0.02476} =161.4252008}[/tex]
[tex]h =\dfrac{161.4252008 \times 0.02476}{ 0.4}[/tex]
h = 9.992 W/m.k
Recall that:
The heat transferred from the plate can be calculate by using the formula:
Heat transferred = h×A ×ΔT
Heat transferred = [tex]h\times A \times (T_2-T_1)[/tex]
Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)
Heat transferred = 22.9 watt
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 pump is to deliver 10, 000 kg/h of toluene at 1140C and 1.1 atm absolute pressure from the Reboiler of a distillation tower to the second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the Reboiler and the pump is 7 kN/m2. The density of toluene is 886 kg/m3. How far above the pump must the liquid be maintained to avoid cavitation
Answer:
3.4093
Explanation:
NPSHa = hatm + hel + hf +hva
the elevation head is the hel
friction loss head is hf
NPSHa is the head of vapour pressure of fluid
atmospheric pressure head is hatm
log₁₀P* = [tex]A -\frac{B}{C+T}[/tex]
[tex]A, B, C are fixed[/tex]
log₁₀Pv = [tex]4.07827-\frac{1343.943}{387.15-53.773}[/tex]
= 4.07827 - 1343.943/333.377
=4.07827 - 4.0313009
= 0.0469691
we take the log
p* = 1.114218
we convert this value to get 111421.8
hvap = 111421.8 * 1/776.14 * 1/9.81
= 14.63
hatm = 1.1 *101325/1 * 1/9.81 *1/776.14
=14.64
hf = 7000/1 * 1/776.14 * 1/9.81
= 0.9193
NPSHa = 2.5
hel = 0.9193 + 2.5 + 14.63 - 14.64
hel = 3.4093
The NSPH values are used to calculate cavitation. The vapor pressure of the liquid is 1.114 atm.
The vapor pressure can be calculated by,
[tex]\mathrm {NPSH_A}= ( \frac {p_i}{\rho g} + \frac {V_i^2}{2g})- \frac {p_v}{\rho g}[/tex]
Where,
[tex]\mathrm {NPSH_A}[/tex] = available NPSH
[tex]p_i[/tex] = absolute pressure at the inlet = 1.1 atm
[tex]V_i[/tex] = average velocity at the inlet = 10, 000 kg/h
[tex]\rho[/tex] = fluid density = 886 kg/m3.
g = acceleration of gravity = 9.8 m/s²
[tex]p_v[/tex] = vapor pressure of the fluid = ?
Put the values in the equation, we get
[tex]p_v = 1.114\ atm[/tex]
Therefore, the vapor pressure of the liquid is 1.114 atm.
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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
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
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 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
The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.
Answer:
μ = 1
F = P√2
Explanation:
The parabola equation is: y = ½ x².
The slope of the tangent is dy/dx = x.
The angle between the tangent and the x-axis is θ = tan⁻¹(x).
At x = 1, θ = 45°.
Draw a free body diagram of the block. There are three forces:
Weight force P pulling down,
Normal force N pushing perpendicular to the surface,
and friction force Nμ pushing up tangential to the surface.
Sum of forces in the perpendicular direction:
∑F = ma
N − P cos 45° = 0
N = P cos 45°
Sum of forces in the tangential direction:
∑F = ma
Nμ − P sin 45° = 0
Nμ = P sin 45°
μ = P sin 45° / N
μ = tan 45°
μ = 1
Draw a new free body diagram. This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.
Sum of forces in the tangential direction:
∑F = ma
F − Nμ − P sin 45° = 0
F = Nμ + P sin 45°
F = (P cos 45°) μ + P sin 45°
F = P√2
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
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.
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
We have that for the Question, it can be said that the average induced emf in the coil is
E=0.028565V
From the question we are told
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Generally the equation for the Average emf induced is mathematically given as
[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]
Hence
[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]
Therefore
[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]
E=0.028565V
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. 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.
A football is kicked with a velocity of 18 m/s at an angle of 20°. What is the
ball's acceleration in the horizontal direction as it flies through the air?
Explanation:
It is given that,
The velocity of football is 18 m/s
It is projected at an angle of 20 degrees
We need to find the ball's acceleration in the horizontal direction as it flies through the air.
When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.
In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.
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
Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <
Answer:
1. Energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared >> visible > ultraviolet
Explanation:
Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.
Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:
1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.
i.e energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.
i.e frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.
i.e wavelength: infrared >> visible > ultraviolet
In terms of lowest to the highest energy,frequency, and wavelength is;
Energy: infrared > visible light > ultraviolet
Frequency: infrared > visible light > ultraviolet
Wavelength: ultraviolet > visible light > infrared
The electromagnetic spectrum:
The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.
The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.Learn more about electromagnetic spectrum here:
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What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
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.