CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
A straight wire that is 0.56 m long is carrying a current of 2.6 A. It is placed in a uniform magnetic field, where it experiences a force of 0.24 N. The wire makes an angle of 900 with the magnetic field. What is the magnitude of the magnetic field
Answer:
0.165TeslaExplanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
Hence the magnitude of the magnetic field is approximately 0.165Tesla
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.
Learn more about Lenz's Law here:
https://brainly.com/question/12876458?referrer=searchResults
What characteristic makes Biology a science, but not Art History?
Using a process of testing ideas and gathering evidence
o Writing books about the subject
O Having a college degree to study it
Discussing and sharing ideas
Answer:
Using a process of testing ideas and gathering evidence
Explanation:
A certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied. In approximately what distance would the car stop had it been going 66.0mph
Answer: 156.02 metre.
Explanation:
Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.
Let us use third equation of motion,
V^2 = U^2 + 2as
Since the car is decelerating, V = 0
And acceleration a will be negative.
U = 33 mph
S = 39 m
Substitute both into the formula
0 = 33^2 - 2 × a × 39
0 = 1089 - 78a
78a = 1089
a = 1089 / 78
a = 13.96 m/h^2
If we assume that the car decelerate at the same rate.
the distance the car will stop had it been going 66.0mph will be achieved by using the same formula
V^2 = U^2 + 2as
0 = 66^2 - 2 × 13.96 × S
4356 = 27.92S
S = 4356 / 27.92
S = 156.02 m
Therefore, the car would stop at
156.02 m
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
A cook preparing a meal for a group of people is an example of
O kinetic energy because he has the ability to make a meal
O potential energy because he has the ability to make a meal
O kinetic energy because he is making the meal
o potential energy because he is making the meal
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
If a vacuum pump reduces the pressure of a gas to 1.0 x 10-6 atm, what is the pressure expressed in millimeters of mercury
Answer:
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
Explanation:
According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:
[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]
[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
A jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in meters per second squared right before taking off? Neglect air resistance and friction.
Answer:
The acceleration is [tex]a =2\ m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the jumbo jet is [tex]m_j = 100000\ kg[/tex]
The thrust is [tex]F_k = 50000 \ N[/tex]
Generally given that the jet has four engines the total thrust is
[tex]F_t = 4 * F_k[/tex]
substituting values
[tex]F_t = 4 * 50000[/tex]
[tex]F_t = 200000 \ N[/tex]
Generally the acceleration of the is mathematically represented as
[tex]a = \frac{F_t}{m}[/tex]
substituting values
[tex]a =2 \frac{N}{kg}[/tex]
Now
[tex]N = kg \cdot m/s^2[/tex]
Hence
[tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]
[tex]a =2\ m/s^2[/tex]
When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?
Answer:
The number of interference fringes is [tex]n = 3[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]
The distance of separation is [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]
The order of maxima is m = 5
The condition for constructive interference is
[tex]d sin \theta = n \lambda[/tex]
=> [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]
=> [tex]\theta = 21.16^o[/tex]
So at
[tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]
[tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]
=> [tex]n = 3[/tex]
Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?
Answer:
29.5°
Explanation:
To find the distance d
d = 1E10^-2/8500lines
= 1.17x 10-6m
But wavelength in first order maximum is 577nm
and M = 1
So
dsin theta= m. Wavelength
Theta= sin^-1 (m wavelength/d)
= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6
= 493*10^-3= sin^-1 0.493
Theta = 29.5°
A diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?
Answer:
(a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Explanation:
Given that,
Width of diffraction grating [tex]w= 19.2\ mm[/tex]
Number of rulings[tex]N=6010[/tex]
Wavelength = 337 nm
We need to calculate the distance between adjacent rulings
Using formula of distance
[tex]d=\dfrac{w}{N}[/tex]
Put the value into the formula
[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]
[tex]d=3.19\times10^{-6}\ m[/tex]
We need to calculate the value of m
Using formula of constructive interference
[tex]d \sin\theta=m\lambda[/tex]
[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]
Here, m = 0,1,2,3,4......
[tex]\lambda[/tex]=wavelength
For largest value of θ
[tex]\dfrac{m\lambda}{d}>1[/tex]
[tex]m>\dfrac{d}{\lambda}[/tex]
Put the value into the formula
[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]
[tex]m>9.46[/tex]
[tex]m = 9[/tex]
(a). We need to calculate the largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=71.9^{\circ}[/tex]
(b). We need to calculate the second largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=57.7^{\circ}[/tex]
(c). We need to calculate the third largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=47.7^{\circ}[/tex]
Hence, (a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Which statement belongs to Dalton’s atomic theory? Atoms have a massive, positively charged center. Atoms cannot be created or destroyed. Atoms can be broken down into smaller pieces. Electrons are located in energy levels outside of the nucleus.
Answer:
the correct statement is
* atoms cannot be created or destroyed
Explanation:
The Datlon atomic model was proposed in 1808 and represents atoms as the smallest indivisible particle of matter, they were the building blocks of matter and are represented by solid spheres.
Based on the previous descriptive, the correct statement is
* atoms cannot be created or destroyed
Answer:
the Answer is b hope it help
Explanation:
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is located 1.03 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
a person Travels along a straight road for half the distance with velocity V1 and the remaining half the distance with velocity V2 the average velocity is given by
Answer: (V1+V2)/2
Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Explanation:
Hey, there!
Here, In question given that,
potential difference (V)= 8V
resistance (R)= 10 ohm
Now,
According to the Ohm's law,
V= R×I { where I = current}
or, I = V/R
or, I = 8/10
Therefore, current is 4/5 A or 0.8 A.
(A= ampere = unit of current).
Hope it helps...
In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?
Answer:
The wavelength is [tex]\lambda = 419 \ nm[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 3.34 *10^{-5} \ m[/tex]
The distance of the screen is [tex]D = 3.30 \ m[/tex]
The order of the fringe is n = 7
The distance of separation of fringes is y = 29.0 cm = 0.29 m
Generally the wavelength of the light is mathematically represented as
[tex]\lambda = \frac{y * d }{ n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]
[tex]\lambda = 4.19*10^{-7}\ m[/tex]
[tex]\lambda = 419 \ nm[/tex]
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B0)(k) extends over a width x=L. The particle is deflected a distance d in the +y direction as it traverses the field. Determine the magnitude of the momentum (p).
Answer:
Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
Explanation:
So, from the question, we are given that the charge = q, the momentum = p.
=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."
Momentum,P = mass × Velocity, v -----(1).
We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.
Radius,r = P × v / B0 -----------------------------(2).
Centrepetal force = q × B0 × v. ----------(3).
(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.
Equating Equation (2) and (3) gives;
P = B0 × q × r.
Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
A typical ten-pound car wheel has a moment of inertia of about 0.35kg *m2. The wheel rotates about the axle at a constant angular speed making 70.0 full revolutions in a time interval of 4.00 seconds. What is the rotational kinetic energy K of the rotating wheel? Express answer in Joules
Answer:
The rotational kinetic energy is [tex]K = 2116.3 \ J[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.35 \ kg \cdot m^2[/tex]
The number of revolution is N = 70 revolution
The time taken is t = 4.0 s
Generally the angular velocity is mathematically represented as
[tex]w = \frac{2 \pi N }{t }[/tex]
substituting values
[tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]
[tex]w = 109.97 \ rad/s[/tex]
The rotational kinetic energy K i mathematically represented as
[tex]K = \frac{1}{ 2} * I * w^2[/tex]
substituting values
[tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]
[tex]K = 2116.3 \ J[/tex]
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W
Complete Question
At what rate must Uranium 235 undergo fission by neutron bombardment to generate energy at a rate of 100 W (1 W = 1 J/s)? Assume each fission reaction releases 200 MeV of energy.
Answer
a. Approximately [tex]5*10^{10}[/tex] fissions per second.
b. Approximately [tex]6*10^{12 }[/tex]fissions per second.
c. Approximately [tex]4*10^{11}[/tex] fissions per second.
d. Approximately [tex]3*10^{12}[/tex] fissions per second.
e. Approximately[tex]3*10^{14}[/tex] fissions per second.
Answer:
The correct option is d
Explanation:
From the question we are told that
The energy released by each fission reaction [tex]E = 200 \ MeV = 200 *10^{6} * 1.60 *10^{-19} =3.2*10^{-11} \ J /fission[/tex]
Thus to generated [tex]100 \ J/s[/tex] i.e (100 W ) the rate of fission is
[tex]k = \frac{100}{3.2 *10^{-11} }[/tex]
[tex]k =3*10^{12} fission\ per \ second[/tex]
When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emf is 12.7 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00458 Wb. How many turns does the solenoid have?
Answer:
[tex]N = 208 \ turns[/tex]
Explanation:
From the question we are told that
The rate of current change is [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]
The magnitude of the induced emf is [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]
The current is [tex]I = 1.50 \ A[/tex]
The average flux is [tex]\phi = 0.00458 \ Wb[/tex]
Generally the number of turns the number of turn the solenoid has is mathematically represented as
[tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]
substituting values
[tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]
[tex]N = 208 \ turns[/tex]
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.
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
[tex]Em_{f}[/tex] = U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h
A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the inductor is
Answer:
The voltage is equal to the batteries terminal voltage
Explanation:
Explanation:
Scouts at a camp shake the rope bridge they have just crossed and observe the wave crests to be 9.70 m apart. If they shake the bridge twice per second, what is the propagation speed of the waves (in m/s)?
Answer:
The speed of the wave is 19.4 m/s
Explanation:
The wave's crest to crest distance (the wavelength of this rope's wave) λ= 9.70 m
The bridge is shaken twice, meaning that two wavelengths passed a given point on the rope per sec. The frequency of a wave is the amount of that wave that passes a given point in a second.
this means that the frequency f = 2 Hz
The speed of a wave = fλ = 9.70 x 2 = 19.4 m/s
A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
(A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Explanation:
Given that,
Magnetic field = 0.45 T
The loop's diameter changes from 17.0 cm to 6.0 cm .
Time = 0.53 sec
(A). We need to find the direction of the induced current.
Using Lenz law
If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.
(B). We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{1}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{1}=0.01021\ Wb[/tex]
We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{2}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{2}=0.00127\ Wb[/tex]
We need to calculate the magnitude of the average induced emf
Using formula of emf
[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]
Put the value into t5he formula
[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]
[tex]\epsilon=0.016867\ V[/tex]
[tex]\epsilon=16.87\ mV[/tex]
(C). If the coil resistance is 2.5 Ω.
We need to calculate the induced current
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{0.016867}{2.5}[/tex]
[tex]I=0.00675\ A[/tex]
[tex]I=6.75\ mA[/tex]
Hence, (A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Given three resistors of different values, how many possible resistance values could be obtained by using one or more of the resistors?
Answer:
8 possible combinations
Assuming R 1, R 2 and R 3 be three different Resistance
1- all three in series
2-all three in parallel
3- R 1 and R 2 in series and parallel with R 3
4-R 1 and R 3 in series and parallel with R 2
5-R 2 and R 3 in series and parallel with R 1
6- R 1 and R 2
in parallel and series with R 3
7-R 1 and R 3 in parallel and series with R 2
8-R 2 and R 3 in V with R 1
