Answer:
Torque = 8.38Nm
Explanation:
Time= 8.00s
angular speed (w) =400 rpm
Moment of inertia (I)= 1.60kg.m2 about its rotation axis
We need to convert the angular speed from rpm to rad/ sec for consistency
2PI/60*n = 0.1047*409 = 41.8876 rad/sec
What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?
Then we need to use the formula below for our torque calculation
from basic equation T = J*dω/dt ...we get
Where : t= time in seconds
W= angular velocity
T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm
Therefore, constant torque that is required is 8.38 Nm
Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.
Given-
Inertia of the flywheel is 1.60 kg m squared.
Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,
[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]
[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]
[tex]\omega = 41.89[/tex]
Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.
When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,
[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]
[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]
[tex]\tau=8.38[/tex]
Hence, the required constant torque is 8.38 N-m.
to know more about the torque, follow the link below-
https://brainly.com/question/6855614
A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answer:
λ = 5.2 x 10⁻⁷ m = 520 nm
Explanation:
From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:
Δx = λL/d
where,
Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m
L = Distance between slits and screen = 3.1 m
d = Separation between slits = 0.0005 m
λ = wavelength of light = ?
Therefore,
0.00322 m = λ(3.1 m)/(0.0005 m)
λ = (0.00322 m)(0.0005 m)/(3.1 m)
λ = 5.2 x 10⁻⁷ m = 520 nm
Recent technological developments like high-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have refined and extended the camera’s capacity to provide information. Which passage assertion does this information support most strongly?
Answer:
D) Photography can be used to both control and benefit society.
Explanation:
High-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have been used to both control and benefits the society in the sense that it has helped to take records of information of crime, traffic offenders such drunk drivers and over speeding drivers, e.t.c. it helps control by given their information and automatically penalizing them or ensuring the agency penalized them and also benefit the society by preventing people from committing crime thereby, protecting them from offenders.
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle
Answer:
T = 2689.6N
Explanation:
Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length. Hence we have :
Torque ( clockwise) = Torque ( anticlockwise)
m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1
Where m1 = 36kg
m2 = 20kg
g = 9.81m/s^2
Theta = 12
Substituting into equation 1
36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)
353.16L/2+196.2L = T ×0.2079(2L/3)
176.58L+196.2L = T × 0.1386L
372.78L = 0.1386LT
T = 372.78L/0.1386L
T = 2689.6N
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?
Answer:
Convection and Radiation mechanisms carry most of the heat
Explanation:
This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.
A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.
Answer:
Explanation:
The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.
In compound microscopes, the distance between the two lenses is expressed as L = v0+ue
v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.
Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.
Given L = 18.0cm
Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm
1/f0 = 1/u0+1/v0
f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.
1/v0 = 1/7.4-1/12
1/v0 = 0.1351-0.0833
1/v0 = 0.0518
v0 = 1/0.2184
v0 = 19.31cm
Note that v0 = ue = 19.31cm
To get ve, we will use the lens formula 1/fe = 1/ue+1/ve
1/ve = 1/fe-1/ue
Given ue = 19.31cm and fe = -7.00cm
1/ve = -1/7.0-1/19.31
1/ve = -0.1429-0.0518
1/ve = -0.1947
ve = 1/-0.1947
ve = -5.14cm
Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens
b) Magnification of the final image M = ve/ue
M = 5.14/19.31
M = 0.27
Magnification of the final image is 0.27
The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct
Answer:
B) the particle's momentum.
Explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
[tex]F_c=\dfrac{m\times V^2}{R}[/tex]
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
[tex]F_m=q\times V\times B[/tex]
At the equilibrium condition
[tex]F_m=F_c[/tex]
[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]
[tex]R=\dfrac{m\times V}{q\times B}[/tex]
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.
Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separated by 0.4 m on a wall 1.62 m from the grating. What is the wavelength of the laser light in nm
Answer:
Wavelength is 4.8x10^-7m
Explanation:
See attached file
What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?
Answer:
The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]
Explanation:
From the question we are told that
The order of maximum diffraction is m = 2
The wavelength is [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]
The angle is [tex]\theta = 62.5^o[/tex]
Generally the condition for constructive interference for diffraction grating is mathematically represented as
[tex]dsin \theta = m * \lambda[/tex]
where d is the distance between the lines on a diffraction grating
So
[tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]
substituting values
[tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]
[tex]d = 1.747 *10^{-6} \ m[/tex]
An optical fiber uses one glass clad with another glass. What is the critical angle? (Assume the glass in the fiber has an index of refraction of 1.69, and the glass in the cladding has an index of refraction of 1.50.)
Answer:
The critical angle is [tex]\theta _c = 62.57^o[/tex]
Explanation:
From the question we are told that
The index of refraction of the glass in the fiber is [tex]n_f = 1.69[/tex]
The index of refraction of the glass in the cladding is [tex]n_c = 1.50[/tex]
The critical angle is mathematically evaluated as
[tex]\theta_c = sin^{-1}[\frac{n_c }{n_f } ][/tex]
substituting values
[tex]\theta_c = sin^{-1}[\frac{1.50 }{1.69 } ][/tex]
[tex]\theta _c = 62.57^o[/tex]
If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
When a current of 2.0 A flows in the 100-turn primary of an ideal transformer, this causes 14 A to flow in the secondary. How many turns are in the secondary
Answer:
14.29 turns.
Explanation:
From the question,
Ns/Np = Ip/Is........................ Equation 1
Where Ns = Secondary turn, Np = Primary turn, Is = current flowing in the secondary turn, Ip = current flowing in the primary turn.
Make Ns the subject of the equation
Ns = NpIp/Is.................... Equation 2
Given: Np = 100 turns, Ip = 2.0 A, Is = 14 A.
Substitute these values into equation 2
Ns = 100(2.0)/14
Ns = 14.29 turns.
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
What portion of the difference in the angular speed before and after you increased the mass can be accounted for by frictional losses
Answer:
As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.
Explanation:
friction can be defined as resistance in motion of bodies in relative to one another
momentum is the product of mass and velocity
torque is the time rate of change in momentum
τ = [tex]\frac{dL}{dt}[/tex]
where L = Iω = mvr
I = moment of inertia
ω= angular frequency
if there is no external force(torque) acting on the system, then
[tex]\frac{dL}{dt}[/tex] = 0
dL = 0 = constant
moment of inertia I depends on the distribution of mass on the axis of rotation.
as the mass increases, the angular momentum(L) increases
angular frequency, ω, remains constant
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
Need help understanding this. If anyone help, that would be greatly appreciated!
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
a) acceleration:
ā=v^2/r
ā=(15m/s)^2/27m
ā=225/27 m/s^2
ā=8.333 m/s^2
force:
F=mā. where the is equal to v^2/r
F=1000kg*8.3 m/s^2
F=8333.3 N
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?
Answer:
Explanation:
let the coil of length l and breathe b entering the magnetic field B with speed v.
So, the magnetic flux through the coil is
Ф = B(l×b)
length × breathe = area
Ф = BA
dФ = BdA
therefore induced emf is given as
ε = [tex]Bl(\frac{db}{dt})[/tex]
note: [tex]\frac{db}{dt} = v[/tex]
ε[tex]= Blv[/tex]
attached is the diagram for the solution
I need help with this question?
Answer:
You got it right, didn't you?
c) vector C
Explanation:
opposite to vector V
Answer:
A, vector B
Explanation:
A negative vector is a vector which points in the opposite direction, even though it’s in a different quadranot it’s still the opposite direction.
Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field
Answer:
case 1 of physics is the answer
You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?
Answer:
I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows
Explanation:
In thermodynamics, coolness can not be transferred, only heat can be transferred
Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.
Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed
Answer:
The number of turns of the wire needed is 243 turns
Explanation:
Given;
self inductance of the solenoid, L = 1.8 mH
cross sectional area of the inductor, A = 1.6 x 10⁻³ m²
length of the inductor, l = 0.066 m
The self inductance of long solenoid is given by;
L = μ₀n²Al
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
n is number of turns per length
A is the area of the solenoid
l is length of the solenoid
[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]
The number of turns is given by;
N = nL
N = (3682.74)(0.066)
N = 243 turns
Therefore, the number of turns of the wire needed is 243 turns
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 77.0 cm behind the grating. First-order maxima are observed at distances of 58.0 cm , 65.4 cm , and 94.5 cm from the central maximum. What are the wavelengths of light emitted by element X?
Answer:
500 nm, 530 nm, 650 nm
Explanation:
Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,
s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm
Now the grating has a slit spacing of -
s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm
The diffraction angles of the " positionings " should thus be -
θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,
θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,
θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°
The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )
Wavelengths : 500 nm, 530 nm, 650 nm
This question will be solved using the "grating equation".
The wavelengths of the light emitted by element X are:
"1. 6.654 x 10⁻⁷ m = 665.4 nm
2. 6.349 x 10⁻⁷ m = 634.9 nm
3. 5.262 x 10⁻⁷ m = 526.2 nm"
The diffraction grating equation is given as follows:
[tex]m\lambda = d Sin\ \theta[/tex]
where,
m = order of maxima = 1
λ = wavelength of light = ?
d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]
d = (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit
θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]
where,
L = distance of grating from the screen = 77 cm
y = distance of maxima from central maxima
Hence, the general equation after substituting constant values becomes:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]
FOR y = 58 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]
λ = 6.654 x 10⁻⁷ m = 665.4 nm
FOR y = 65.4 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]
λ = 6.349 x 10⁻⁷ m = 634.9 nm
FOR y = 94.5 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]
λ = 5.262 x 10⁻⁷ m = 526.2 nm
The attached picture shows the arrangement of the light rays in a diffraction grating.
Learn more about diffraction grating here:
https://brainly.com/question/17012571?referrer=searchResults
A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?
Answer:
The magnetic field is [tex]B = 0.0764 \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal is [tex]m = 0.210 \ kg[/tex]
The current is [tex]I = 11.0 \ A[/tex]
The distance between the rail(length of the rod ) is [tex]d = 0.490 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.200[/tex]
Generally the magnetic force is mathematically represented as
[tex]F_b = B * I * d[/tex]
Given that the rod is moving at a constant velocity, it
=> [tex]F_b = F_k[/tex]
Where [tex]F_k[/tex] is the kinetic frictional force which is mathematically represented as
[tex]F_k = \mu_k * m * g[/tex]
So
[tex]B * I * d = \mu_k * m * g[/tex]
=> [tex]B = \frac{\mu_k * m * g}{I * d }[/tex]
substituting values
=> [tex]B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }[/tex]
=> [tex]B = 0.0764 \ T[/tex]
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron
Answer:
The speed of the electron is 6.79 x 10⁵ m/s
The radius of the circular path is 1.357 x 10⁻⁵ m
Explanation:
Given;
magnetic field, B = 0.285 T
energy of electron, E = 2.10 x 10⁻¹⁹ J
The kinetic energy of the electron is calculated as;
[tex]K.E = \frac{1}{2} m_eV^2[/tex]
Where;
[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg
V is the speed of the electron
[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]
The radius of the circular path is given by;
[tex]R = \frac{M_eV}{qB}[/tex]
where;
q is the charge of the electron = 1.6 x 10⁻¹⁹ C
[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]
greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.
Answer:
[similar to]
Explanation:
it is the missing word
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.