Answer:
genus yds it's the
Explanation:
xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj
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]
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
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) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle
tysm to u who answers :)
1. The ocean water collects back in the ocean.
2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
3. an excessive amount of water flowing from downslope along earths surface
4. A.Evaporation occurs when water is warmed by the sun.
5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.
Hope this answer correct ✌️
Specific heat is a measurement of the amount of heat energy input required for one gram of a substance to increase its temperature by one degree Celsius. Solid lithium has a specific heat of 3.5 J/g·°C. This means that one gram of lithium requires 3.5 J of heat to increase 1°C. Plot the temperature of 1g of lithium after 3.5, 7, and 10.5 J of thermal energy are added.
Answer:
ΔT = 1ºC , 2ºCand 3ºC
Explanation:
In this exercise they indicate the specific heat of lithium
let's calculate the temperature increase as a function of the heat introduced
Q = m [tex]c_{e}[/tex] ΔT
ΔT = Q / m c_{e}
calculate
for Q = 3.5 J
ΔT = 3.5 / (1 3.5)
ΔT = 1ºC
For Q = 7.0 J
ΔT = 7 / (1 3.5)
ΔT = 2ºC
for Q = 10.5 J
ΔD = 10.5 / (1 3.5)
ΔT = 3ºC
we see that this is a straight line, see attached
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
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answer:
C $0.75 my friend I wish it is right answer
A wire of 5.8m long, 2mm diameter carries 750ma current when 22mv potential difference is applied at its ends. if drift speed of electrons is found then:_________.
(a) The resistance R of the wire(b) The resistivity p, and(c) The number n of free electrons per unit volume.
Explanation:
According to Ohms Law :
V = I * R
(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms
Also,
[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]
(B)
[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]
Drift speed is missing. It is given as;
1.7 × 10^(-5) m/s
A) R = 0.0293 ohms
B) ρ = 1.589 × 10^(-8)
C) n = 8.8 × 10^(28) electrons
This is about finding, resistance and resistivity.
We are given;Length; L = 5.8 m
Diameter; d = 2mm = 0.002 m
Radius; r = d/2 = 0.001 m
Voltage; V = 22 mv = 0.022 V
Current; I = 750 mA = 0.75 A
Area; A = πr² = 0.001²π
Drift speed; v_d = 1.7 × 10^(-5) m/s
A) Formula for resistance is;R = V/I
R = 0.022/0.75
R = 0.0293 ohms
B) formula for resistivity is given by;ρ = RA/L
ρ = (0.0293 × 0.001²π)/5.8
ρ = 1.589 × 10^(-8)
C) Formula for current density is given by;J = n•e•v_d
Where;
J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²
e is charge on an electron = 1.6 × 10^(-19) C
v_d = 1.7 × 10^(-5) m/s
n is number of free electrons per unit volume
Thus;
238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))
238732.44 = (2.72 × 10^(-24))n
n = 238732.44/(2.72 × 10^(-24))
n = 8.8 × 10^(28)
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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?).
All household circuits are wired in parallel. A 1140-W toaster, a 270-W blender, and a 80-W lamp are plugged into the same outlet. (The three devices are in parallel when plugged into the same outlet.) Assume that this is the standard household 120-V circuit with a 15-A fuse.
a. What current is drawn by each device?
b. To see if this combination will blow the 15-A fuse, find the total current used when all three appliances are on.
Answer:
total current = 12.417 A
so it will not fuse as current is less than 15 A
Explanation:
given data
toaster = 1140-W
blender = 270-W
lamp = 80-W
voltage = 120 V
solution
we know that current is express as
current = power ÷ voltage ......................1
here voltage is same in all three device
so
current by toaster is
I = [tex]\frac{1140}{120}[/tex]
I = 9.5 A
and
current by blender
I = [tex]\frac{270}{120}[/tex]
I = 2.25 A
and
current by lamp is
I = [tex]\frac{80}{120}[/tex]
I = 0.667 A
so here device in parallel so
total current is = 9.5 A + 2.25 A + 0.667 A
total current = 12.417 A
so it will not fuse as current is less than 15 A
A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .Part A. Calcualte the coil's self-inductance.Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf froma to b or from b to a?
Complete Question
A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .
Part A. Calculate the coil's self-inductance.
Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.
Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?
Answer:
Part A
[tex]L = 0.000863 \ H[/tex]
Part B
[tex]\epsilon = 0.863 \ V[/tex]
Part C
From terminal a to terminal b
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The cross-sectional area is [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]
The radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]
Generally the coils self -inductance is mathematically represented as
[tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]
[tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]
[tex]L = 0.000863 \ H[/tex]
Considering the Part B
Initial current is [tex]I_1 = 5.00 \ A[/tex]
Current at time t is [tex]I_t = 3.0 \ A[/tex]
The time taken is [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]
The self-induced emf is mathematically evaluated as
[tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]
=> [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]
substituting values
[tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]
[tex]\epsilon = 0.863 \ V[/tex]
The direction of the induced emf is from a to b because according to Lenz's law the induced emf moves in the same direction as the current
This question involves the concepts of the self-inductance, induced emf, and Lenz's Law
A. The coil's self-inductance is "0.863 mH".
B. The self-induced emf in the coil is "0.58 volts".
C. The direction of the induced emf is "from b to a".
A.
The self-inductance of the coil is given by the following formula:
[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]
where,
L = self-inductance = ?
[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 590
A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²
r = radius = 5 cm = 0.05 m
Therefore,
[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]
L = 0.863 x 10⁻³ H = 0.863 mH
B.
The self-induced emf is given by the following formula:
[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]
where,
E = self-induced emf = ?
ΔI = change in current = 2 A
Δt = change in time = 3 ms = 0.003 s
Therefore,
[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]
E = 0.58 volts
C.
According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.
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light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
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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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
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answer:
1.125×10⁻⁹ J
Explanation:
Applying,
E = 1/2CV²................... Equation 1
Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.
Given; C = 1.0 nF, = 1.0×10⁻⁹ F, V = 1.5 V
Substitute into equation 1
E = 1/2(1.0×10⁻⁹×1.5²)
E = 1.125×10⁻⁹ J
Hence the energy stored by the capacitor is 1.125×10⁻⁹ J
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
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:
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
Explanation:
A) Vair = 1.3 L
B) Volume is not reasonable
Explanation:
A)
Assume
m to be total mass of the man
mp be the mass of the man that pulled out of the water
m1 be the mass above the water with the empty lung
m2 be the mass above the water with full lung
wp be the weight that the buoyant force opposes as a result of the air.
Va be the volume of air inside man's lungs
Fb be the buoyant force due to the air in the lung
given;
m = 78.5 kg
m1 = 3.2% × 78.5 = 2.5 kg
m2 = 4.85% × 78.5 = 3.8kg
But, mp = m2- m1
mp = 3.8 - 2.5
mp = 1.3kg
So using
Archimedes principle, the relation for formula for buoyant force as;
Fb = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
Fb = wp = 1.3× 9.81
Fb = 12.7N
But
Fb = (ρ_water × V_air × g)
So
Vair = Fb/(ρ_water × × g)
Vair = 12.7/(1000 × 9.81)
V_air = 1.3 × 10^(-3) m³
convert to litres
1 m³ = 1000 L
Thus;
V_air = 1.3× 10^(-3) × 1000
V_air = 1.3 L
But since the average lung capacity of an adult human being is about 6-7litres of air.
Thus, the calculated lung volume is not reasonable
Explanation:
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
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
UVC light used in sterilizers, has wavelengths between 100 to 280 nm. If a certain UVC wave has a wavelength of 142.9 nm, what is the energy of one of its photons in J
Answer:
The energy of one of its photons is 1.391 x 10⁻¹⁸ J
Explanation:
Given;
wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m
The energy of one photon of the UVC light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f is frequency of the light
f = c / λ
where;
c is speed of light = 3 x 10⁸ m/s
λ is wavelength
substitute in the value of f into the main equation;
E = hf
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]
Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ 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
How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?
Answer
1.07E22 Joules
Explanation;
We know that mass expands by a factor
=>>1/√[1-(v/c)²]
But v= 0.509c
So
1/√(1 - 0.509²)
=>>> 1/√(1 - 0.2591)
= >> 1/√(0.7409) = 1.16
But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy
And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.
So Multiplying by 119.84
Kinetic energy will be 1.07x 10^22 joules
In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.
Answer:
a. to the west.
Explanation:
An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.
A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
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