Answers
Explanation:
ur answer is in attachment.
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Explanation:
Force applied on the box = 20 N
Displacement, s = 10 m
Time taken = 5 sec
According to first condition of the question, we could find the value of work done
i.e
Work done = force × displacement
= 20 × 10
= 200 Joule
According to second condition of the question, we could find the value of power
i.e
Power = work done/Time taken
= 200/5
= 40 watt.
Hope it helpful!!!!!!!!!!!!
Related Questions
The speed of a boat is often given in knots. If a speed of 5 knots were expressed in the SI system of units, the units would be:____________.
Answers
Answer:
0.514 m/s
Explanation:
The knot is a unit of nautical speed used in maritime navigation.
1 knot is equal to one nautical mile per hour,
1 nautical mile per hour = 1.852 km/h
The basic SI system of units of speed is in 'm/s'
1.852 km/h = (1.852 x 1000)/3600 = 0.514 m/s
The elements within a group tend to share
A. similar chemical properties and characteristics
B. similar atomic weights
c. similar atomic numbers
D. similar atomic symbols
Answers
Answer: A. similar chemical properties and characteristics
What is the maximum number of electrons in an atom that can have the following quantum numbers?
a. n = 3, m1 = -2;
b. n = 4, l = 3;
c. n = 5, l = 3, ml = 2;
d. n = 4, l = 1 ml = 0.
Answers
Answer:
Explanation:
a ) n = 3, m1 = -2
when n = 3 , l = 0 , 1 , 2 .
when l = 2 , one electron may have m = -2 .
two electrons with s = + 1/2 and - 1/2 may have m = -2 .
So only two electrons may have the configuration of n = 3 , m = -2 .
b ) n = 4, l = 3
when , n = 4 and l = 3 , m may have values -3 , -2 , -1 , 0 , 1 , 2 , 3 . Each m have two electrons with s = + 1/2 and - 1/2 .
So altogether there are 7 x 2 = 14 electrons that have configuration of n = 4, l = 3 .
c ) n = 5, l = 3, ml = 2
Similar to case of a ) , only two electrons with s = + 1/2 and - 1/2 may have configuration of n = 5, l = 3, ml = 2
d ) n = 4, l = 1 ml = 0.
In this case also only two electrons may have configuration of
n = 4, l = 1 ml = 0. with s = + 1/2 and - 1/2 .
(a) The maximum number of electrons in an atom that have n = 3, ml = -2 is two electrons.
(b) The maximum number of electrons in an atom that have n = 3, ml = -2 is 14 electrons.
(c) The maximum number of electrons in an atom that have n = 5 l = 3, ml = 2 is two electrons.
(d) The maximum number of electrons in an atom that have n = 4 l = 1 , ml = 0 is two electrons.
The given parameters:
n = 3, ml = -2n = 4, l = 3n = 5, I = 3, ml = 2n = 4, I = ml = 0The maximum number of electrons in an atom that have n = 3, ml = -2
the quantum number = 3the maximum number of electron = 2The maximum number of electrons in an atom that have n = 4, l = 3
the quantum number = 4the maximum number of electron = (2 x 3) + 1 = 7The orbital is 7 which corresponds to f-orbital and the maximum number of electrons = 14;
The maximum number of electrons in an atom that have n = 5 l = 3, ml = 2
the quantum number = 5the maximum number of electrons = 2The maximum number of electrons in an atom that have n = 4 l = 1 , ml = 0
the quantum number = 4the maximum number of electrons = 2Learn more about quantum number of electrons here: https://brainly.com/question/11575590
PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large container of water with a higher temperature. Which choice correctly explains what happens to the thermal energy of these systems? A)The smaller container of water has more thermal energy than the larger container, and some of that energy is transferred to the warmer water in a process known as heating. B)The larger container of water has more thermal energy and some of that energy is transferred to the colder water in a process known as heating. C)The larger container of water has more heat and thermal energy than the smaller container. Some heat and thermal energy is transferred to the smaller container of water. D)The smaller container of water has more heat and thermal energy than the larger container. Some heat and thermal energy is transferred to the larger container of water.
Answers
Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce ([tex]T_{f}[/tex]-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
A 76-W incandescent light bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?
Answers
Answer:
43200c
Explanation:
Relative to the frame of the observer making the measurement, at what speed parallel to its length is the length of a meterstick 60 cm?
Answers
Answer:
The speed of the observer is 2.4 x 10^8 m/s
Explanation:
The standard length of a meter stick is 100 cm
we are to calculate at what speed parallel to the length the length reduces to 60 cm.
This is a relativistic effect question. We know that the length will contract to this 60 cm following the equation below
[tex]l = l_{0}\sqrt{1 - \beta ^{2} }[/tex]
where
[tex]l[/tex] is the new length of 60 cm
[tex]l_{0}[/tex] is the original length which is 100 cm
[tex]\beta[/tex] is the the ratio v/c
where
v is the speed of the observer
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
60 = [tex]100\sqrt{1 - \beta ^{2} }[/tex]
0.6 = [tex]\sqrt{1 - \beta ^{2} }[/tex]
we square both side
0.36 = 1 - [tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 1 - 0.36 = 0.64
β = [tex]\sqrt{0.64}[/tex] = 0.8
but β = v/c
v/c = 0.8
substituting value of c, we have
v = 0.8 x 3 x 10^8 = 2.4 x 10^8 m/s
Is it possible for a radioactive nucleus to decay two times and end up as the same element as the original?
Answers
Answer:
Yes, it is possible for a radioactive nucleus to decay two times and end up as the same element as the original
Explanation:
Radioactive decay of nucleus is the process by which an unstable atomic nucleus loses energy by radiation. The energy loss may be from an excited nucleus, which may be emitted as a gamma ray in a process called gamma decay or due to a mass deficit whenever the nucleus decays according to the energy-mass relationship. Whenever a radioactive nucleus decays, new nucleus may be formed in a "transmutation decay", usually into an element of a lower atomic mass than the mother element. Another type of radioactive decay results in products that vary, appearing as two or more "fragments" of the original nucleus with a range of possible masses. This decay is called "spontaneous fission" and it usually happens when a large unstable nucleus decays into two (or occasionally three) smaller daughter nuclei, with an emission of gamma rays, neutrons, or other particles.
Convert 50km/hr into m/s using dimensional analysis?
Answers
Answer:
Explanation:
1 km=1000 m
1 hour =60 min =60*60 sec=3600 sec
Now put 1000 m instead of km and 3600 sec instead of hour in the given expression.
=50 km/hour
=50*1000 m/3600 sec
=500 m/36 sec
=13.89 m/s
Answer:
= 13.89 m/s
Explanation:
km 1000 m 1 hr
50 ----- x ---------- x ------------
hr 1 km 3600 secs
= 13.89 m/s
A nearsighted woman has a far point of 180 cm . Part A. What kind of lens, converging or diverging, should be prescribed for her to see distant objects more clearly?Part B. What power should the lens have?
Answers
Explanation:
It is given that,
A nearsighted woman has a far point of 180 cm.
The object distance for nearsightedness is infinity, u = ∞
The image distance is, v = -180 cm
Using lens formula,
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-180)}-\dfrac{1}{\infty}\\\\f=-180\ cm\\\\f=-1.8\ m[/tex]
For nearsightedness, a diverging lens is used.
Power of the lens,
P = 1/f
[tex]P=\dfrac{1}{-1.8}\\\\P=-0.56\ D[/tex]
So, the power of the lens is -0.56 D.
The power of lens to be prescribed is 3.5 D.
A person that is far sighted can see far objects clearly but not nearby objects. The farpoint of a normal eye is infinity. Far sightedness is corrected by the use of a convex lens. The near point of the normal eye is 25 cm.
We have that;
u = 25 cm
v = -180 cm
1/f = 1/u - 1/v
1/f = 1/25 - 1/180
1/f = 0.04 - 0.0056
f = 29 cm
Power of lens = 100/f = 100/29 = 3.5 D
Learn more about lens: https://brainly.com/question/1195122
A refracting telescope has a 1.48 m diameter objective lens with focal length 15.4 m and an eyepiece with focal length 3.28 cm. What is the angular magnification of the telescope
Answers
Answer:
m = 469.51
Explanation:
Given that,
The diameter of a refracting telescope is 1.48 m
Focal length of the objective lens is 15.4 m
Focal length of an eyepiece is 3.28 cm
We need to find the angular magnification of the telescope. The ratio of focal length of objective lens to the focal length of the eye piece is called angular magnification of the telescope. So,
[tex]m=\dfrac{f_o}{f_e}\\\\m=\dfrac{15.4}{3.28\times 10^{-2}}\\\\m=469.51[/tex]
So, the angular magnification of the telescope is 469.51.
The uncertainty in position of a proton confined to the nucleus of an atom is roughly the diameter of the nucleus. If this diameter is d = 7.8×10−15 m, what is the uncertainty in the proton's momentum?
Answers
Answer:
0.135E-19kgm/s
Explanation:
Using the uncertainty principle, we find
Dp = h / (4π Dx)
= (6.63×10-34Js)/4π(3.90×10-15m) = 0.135×10-19kg m/s
Three identical point charges, Q = 2 PC, are placed at the vertices of an equilateral triangle as shown in the figure. The length of each side of the triangle is d
= 2 m. Determine the magnitude and direction of the total electrostatic force on the charge at the top of the triangle.
Answers
Answer:
The total force on the charge at the top of the triangle is approximately
[tex]1.56\,\,10^{-14}\,\,N[/tex]
Explanation:
Please look at the attached image to follow the explanation.
Since all charges are of the same positive value, the force exerted on the top charge by the other two are going to be on the line that joins the top charge with each of the other vertices of the triangle, pointing away from the top charge (illustrated by green vectors of the same length in the image).
We need to find the x and y components of these force vectors in order to find the resultant force by combining the x components among themselves, and the y components among themselves. Notice that the angle needed is in all cases [tex]60^o[/tex].
The x- components include the cosine of [tex]60^o[/tex], while the y components include the sine of [tex]60^o[/tex].
Notice as well that the x-components cancel each other (they have the same magnitude but point in opposite directions), while the y-components are both of the same value but pointing in the same direction (pointing up).
Then we just need to multiply by two the y component of one of the forces to find this total force.
Now, the magnitude of the forces in question are given by Coulomb's Law:
[tex]F_C=k\,\frac{q_1\,\,q_2}{d^2} =9\,\,10^9\,\frac{2\,\,10^{-12}\,\,2\,\,10^{-12}}{2^2} =9\,\,10^{-15}\,\,N[/tex]
therefore we can calculate what the y-component of this is using:
[tex]F_y=F\,sin(60^o)=9\,\,10^{-15} \,\frac{\sqrt{3} }{2} \\[/tex]
and twice this value becomes:
[tex]2\, \,F_y=(2)\,\,9\,\,10^{-15} \,\frac{\sqrt{3} }{2}=9\,\,\sqrt{3} \,\,10^{-15}\approx 1.56\,\,10^{-14}\,\,N[/tex]
(The figure model is attached below in a picture) For the figure above there is no friction between block A and the ramp. The mass of block A is 5.6kg. The ramp is 37.383198° from horizontal. Block A is being pulled up the ramp with a constant velocity by a line that is parallel to the ramp. What is the mass of block B?
Give a variable legend for this problem. The model for this problem:
mB (Mass of block B) =__________________________________ Answer________________________________
Answers
Tension, normal reaction and sliding by gravity.
Forces on Block B:
Tension, and weight.
Here it is assumed that the string is light (negligible mass) and inextensible. Hence tension is equal on both blocks. Assumption 2: Block B moves down, so down is considered positive.
Sum of forces: (Block A)
T - mg sin theta = ma. ——— 1
Sum of forces: (Block B)
mg - T = ma. ——— 2
Acceleration is equal for both equations.
Adding 1 and 2,
mg - T + T - mg sin theta = (Total mass)a.
Let the mass of block B be x.
——> xg - 5.6g sin 37.383198 = (5.6 + x)a
——> 10x - 56 sin 37.383198 = (5.6 + x)a (Taking g as 10 m/s^2)
——> 10x - 56*3/5 = (5.6+x)a (Sin 37 can be taken as 3/5 from a triangle of sides 3,4,5)
——> 10x - 33.6 = (5.6+x)a
——> 10x - 33.6 = 0. (Acceleration is 0 as velocity is constant)
x = 3.6kg.
If three forces are equal and their resultant force is zero, then measure of its internal angle is: a)0° b)90° c)60° d)45° Fill in the blanks, tell me which one is right. a or b or c or d
Answers
Answer:
edqr3fr =wdmmwfmnl[=kvlfkgvb
Explanation:bcz idk
Me BeLIves iT iS ZeRo So A
Q2: W A soccer ball is kicked off the ground at an angle
of 40 degrees at a speed of 40mls.
co Find The maximum height that the ball will reach
Answers
Answer:
1metre
Explanation:
height gained is equal to speed multiplied by angle
An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 43.0 km . To the dismay of scientists on earth, an electrical fault causes an on-board thruster to fire, decreasing the speed of the spacecraft by 23.0 m/s .If nothing is done to correct its orbit, with what speed (in km/h) will the spacecraft crash into the lunar surface?
Answers
Answer: v₂ = 5962 km
the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
Explanation:
Given that;
Lunar surface is in an altitude h = 43.0 km = 43 × 10³ m
we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²
speed of the space craft when it crashes into the lunar surface , v
decreasing speed of the space craft = 23 m/s
Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r
but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²
HERE r = Rm + h
we substitute
r = 1.74 × 10⁶ m + 43 × 10³ m
= 1.783 × 10⁶ m
On equating these, we have
G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)
v²/r = GM/r²
v = √ ( GM/r)
v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )
v = √ (2750787.9978)
v = 1658.55 m/s
Now since speed is decreasing by 23 m/s
the speed of the space craft into the lunar face is,
v₁ = 1658.55 m/s - 23 m/s
v₁ = 1635.55 m/s
Now applying conversation of energy, we say
1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)
v₂ = √ [ v₁² + GMe (1/Rm - 1/r)]
v₂ = √ [ 1635.55² + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]
v₂ = √ (2675023.8025 + 67979.24)
v₂ = √(2743003.046)
v₂ = 1656.2 m/s
now convert
v₂ = 1656.2 × 1km/1000m × 3600s/1hrs
v₂ = 5962 km
Therefore the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
The electric field in a traveling em wave in the vacuum has an rms intensity of 6.75 v/m.
Required:
Calculate the amount of energy delivered in average during 3.02x10^7s to 2.00 cm^2 of a wall that it hits perpendicularly?
Answers
Answer:
The amount of energy delivered is 730.84 J
Explanation:
Given;
rms intensity of the electric field, [tex]E_{rms}[/tex] = 6.75 v/m
area of the wall, A = 2.0 cm² = 2.0 x 10⁻⁴ m²
The average intensity of the wave is given by;
[tex]I_{avg} = c \epsilon_o E_{rms}^2[/tex]
where;
c is the speed of light
ε₀ is permittivity of free space
I = (3 x 10⁸)(8.85 x 10⁻¹²)(6.75)²
I = 0.121 W/m²
Average power delivered, P = I x A
P = 0.121 x 2 x 10⁻⁴
P = 2.42 x 10⁻⁵ W
The amount of energy delivered is calculated as;
E = P x t
E = 2.42 x 10⁻⁵ x 3.02 x 10⁷
E = 730.84 J
Therefore, the amount of energy delivered is 730.84 J
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answers
Answer:
The relationship is [tex]E=\frac{ kQ}{d^2}[/tex]
Explanation:
The electric field strength is denoted by the symbol E,
the test charge is denoted be q and the source charge be Q
distance is denoted by d
Then the equation can be rewritten in symbolic form as
Electric field strength is = Force/charge
[tex]E=\frac{F}{q}------1[/tex]
we know that the formula for force is given as
[tex]F= \frac{kQq}{d^2} ------2[/tex]
where [tex]k= 9*10^9 N.m^2/C^2[/tex]
and d is the separation distance between charges
We can insert the expression for Force in equation one
we have
[tex]E= \frac{kQ\sout{q}/d^2}{\xout{q}}--------3[/tex]
We can strike out both qs in the numerator and denominator we have
[tex]E=\frac{ kQ}{d^2}[/tex]
You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a 44 cm radius of curvature.
(a) Where is the image of your 5.0-cm-tall nose located?
(b) What are the image’s size, orientation, and nature (real or virtual)?
Answers
Answer:
a) q = 39.29 cm , b) h ’= - 3.929 cm the image is inverted and REAL
Explanation:
For this exercise we will use the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length of the salad bowl, p and q are the distance to the object and the image
The metal salad bowl behaves like a mirror, so its focal length is
f = R / 2
f = 44/2
f = 22 cm
a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image
1 / q = 1 / f - 1 / p
1 / q = 1/22 - 1/50
1 / q = 0.0254
q = 39.29 cm
b) to calculate the size of the image we use the equation of magnification
m = h’/ h = - q / p
h ’= - q / p h
h ’= - 39.29 / 50 5
h ’= - 3.929 cm
the negative sign means that the image is inverted
as the rays of light pass through the image this is REAL
A beam of light in air strikes a slab of glass (n = 1.52) and is partially reflected and partially refracted and partially refracted, find the angle of incidence if the angle of reflection is twice the angle of refraction.
θ1= (answer in degress)
Answers
Answer:
Explanation:
angle of reflection = angle of incidence i = 2θ
angle of refraction r = θ
sin i / sin r = 1.52
sin2θ / sin θ = 1.52
2 sinθ . cosθ / sin θ = 1.52
2 cosθ = 1.52
cosθ = .76
θ= 41°
Hence angle of incidence = 41°
.
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s. Part A What is the electric field strength inside the solenoid at a point on the axis
Answers
Complete Question
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s.
Part A What is the electric field strength inside the solenoid at a point on the axis?
Part B
What is the electric field strength inside the solenoid at a point 1.50 cm from the axis?
Answer:
Part A
[tex]E = 0 \ V/m[/tex]
Part B
[tex]E_{15} = 0.0345 \ V/m[/tex]
Explanation:
From the question we are told that
The diameter of the solenoid is [tex]d = 5.0 \ cm = 0.05 \ m[/tex]
The magnetic field is [tex]B = 2.0 \ T[/tex]
The rate of the change of the magnetic field is [tex]\frac{dB}{dt} = 4.60 \ T/s[/tex]
The radius of the solenoid is mathematically represented as
[tex]R = \frac{ d}{2}[/tex]
substituting values
[tex]R = \frac{ 5.0 *10^{-2}}{2} = 0.025 \ m[/tex]
Generally the of the solenoid is mathematically represented as
[tex]E = \frac{ r}{2} * |\frac{dB}{dt} |[/tex]
Now at the point on axis is r = 0 given that the axis is the origin so
[tex]E = \frac{ 0}{2} * |\frac{dB}{dt} |[/tex]
[tex]E = 0 \ V/m[/tex]
Now the electric field strength inside the solenoid at a point 1.50cm from the axis is mathematically represented as
[tex]E_{15} = \frac{ 15*10^{-2 }}{2} * |4.60 |[/tex]
[tex]E_{15} = 0.0345 \ V/m[/tex]
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Answers
Answer:
[tex]f=1.98\times 10^5\ Hz[/tex]
Explanation:
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by :
[tex]B=1.2\times 10^{-6}\sin [2\pi (\dfrac{z}{240}-\dfrac{10^7t}{8})][/tex] .....(1)
The general equation of the magnetic field wave is given by :
[tex]B=B_o\sin (kz-\omega t)[/tex] ....(2)
Equation (1) is in form of equation (2), if we compare equation (1) and (2) we find that,
[tex]\omega=\dfrac{10^7}{8}[/tex]
We need to find the frequency of the wave. It is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{10^7}{8\times 2\pi}\\\\f=1.98\times 10^5\ Hz[/tex]
So, the frequency of the wave is [tex]1.98\times 10^5\ Hz[/tex]
A 12.0-g sample of carbon from living matter decays at the rate of 162.5 decays/minute due to the radioactive 14C in it. What will be the decay rate of this sample in 1000 years? What will be the decay rate of this sample in 50000 years?
Answers
Answer:
a)143.8 decays/minute
b)0.41 decays/minute
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of C-14= 5670 years
t= time taken to decay
Ao= activity of a living sample
A= activity of the sample under study
a)
0.693/5670 = 2.303/1000 log(162.5/A)
1.22×10^-4 = 2.303×10^-3 log(162.5/A)
1.22×10^-4/2.303×10^-3 = log(162.5/A)
0.53 × 10^-1 = log(162.5/A)
5.3 × 10^-2 = log(162.5/A)
162.5/A = Antilog (5.3 × 10^-2 )
A= 162.5/1.13
A= 143.8 decays/minute
b)
0.693/5670 = 2.303/50000 log(162.5/A)
1.22×10^-4 = 4.61×10^-5 log(162.5/A)
1.22×10^-4/4.61×10^-5 = log(162.5/A)
0.26 × 10^1 = log(162.5/A)
2.6= log(162.5/A)
162.5/A = Antilog (2.6 )
A= 162.5/398.1
A= 0.41 decays/minute
A skateboarder lands on all four wheels after riding a railing. If the skateboarder has a weight of 990 and the area on the bottom of a single wheel is 0.0005m^2, what pressure does the skateboard put on the ground?
Answers
Explanation:
Pressure = force / area
P = 990 N / (4 × 0.0005 m²)
P = 495,000 Pa
A coiled telephone cord forms a spiral with 52.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.5 cm. Determine the inductance of one conductor in the unstretched cord.
Answers
Answer:
The inductance of one conductor in the unstretched cord is 0.7849 μH
Explanation:
Given;
number of turns of the coil, N = 52 turns
diameter of the coil, d = 1.30 cm
radius of the coil, r = d/2 = 1.30 cm / 2 = 0.65 cm = 0.0065 m
length of the unstretched cord, l = 57.5 cm = 0.575 m
The inductance of one conductor in the unstretched cord is given by;
[tex]L = \frac{N^2 \mu_o A}{l}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
A is area of coil, = πr² = π x (0.0065)² = 1.328 x 10⁻⁴ m²
[tex]L = \frac{(52)^2(4\pi *10^{-7})(1.328*10^{-4})}{0.575} \\\\L = 7.849*10^{-7} \ H[/tex]
L = 0.7849 μH
Therefore, the inductance of one conductor in the unstretched cord is 0.7849 μH
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?
Answers
Answer:
Option C. 70 Ω
Explanation:
Data obtained from the question include:
Resistor (R) = 20 Ω
From diagram given ABOVE, we observed the following
1. R and R are in parallel connections.
2. 2R and 2R are in parallel connections.
3. 4R and 4R are in parallel connections.
Next, we shall determine the equivalent resistance in each case.
This is illustrated below:
1. Determination of the equivalent resistance for R and R parallel connections.
R = 20 Ω
Equivalent R = (R×R) /(R+R)
Equivalent R = (20 × 20) /(20 + 20)
Equivalent R = 400/40
Equivalent R = 10 Ω
2. Determination of the equivalent resistance for 2R and 2R parallel connections.
R = 20 Ω
2R = 2 × 20 = 40 Ω
Equivalent 2R = (2R×2R) /(2R+2R)
Equivalent 2R = (40 × 40) /(40 + 40)
Equivalent 2R = 1600/80
Equivalent 2R = 20 Ω
3. Determination of the equivalent resistance for 4R and 4R parallel connections.
R = 20 Ω
4R = 4 × 20 = 80 Ω
Equivalent 4R = (4R×4R) /(4R+4R)
Equivalent 4R = (80 × 80) /(80 + 80)
Equivalent 4R = 6400/160
Equivalent 4R = 40 Ω
Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:
Equivalent of R = 10 Ω
Equivalent of 2R = 20 Ω
Equivalent of 4R = 40 Ω
Equivalent =?
Equivalent = Equivalent of (R + 2R + 4R)
Equivalent = 10 + 20 + 40
Equivalent = 70 Ω
Therefore, the equivalent resistance between point A and B is 70 Ω.
Refraction in ocean waves is identical to refraction in sound and energy waves in that it involves movement through a different medium.a. Trueb. False
Answers
Answer:
True
Explanation:
Refraction is the change in the speed, wavelength and direction of a wave as it crosses the boundary between two different media of different densities.
Hence, refraction always involves the movement of a wave from one medium to another. This is the key point to be remembered whether we are discussing refraction in ocean waves or sound waves.
Which statement describes a force acting on an object?
A.a person sees a rolling ball toward him
B. A magnet pulls a nail toward it
C. A piece of charcoal burns in a grill.
D.A car is parked on the street
Answers
Answer: B. A magnet pulls a nail towards it
Explanation: Out of the list of answers, this is the only answer where a force is acting on an object. the act of the magnet pulling on the nail is the force, and the nail is the object.
Suppose the moon was twice as a massive as it is. Would the month be Select one: a. half as long bthe same c. twice as long d. 1/4 as long e. 4 times as long
Answers
Answer:
The correct answer is B
Explanation:
The lunar month is the time it takes for the moon to go around the Earth, for this we use Newton's second law where the force is the universal force of attraction
F = ma
the universal attractive force is
F = G M_earth m_moon / r²
Accelerations centripetal acceleration
a = w² r
the angular velocity is for the movement is
w = 2π / T
Where T is the period of revolution; let's substitute
G M_eath m_moon / r² = m_moon (2π /T)² r
G M_earth = 4π² / T² r³
T² = (4π² / G M_eart) r³
Let's analyze this equation we see that it does not depend on the mass of the Luma, therefore the period is the same
The correct answer is B
If the speed of a car is 20m/s .How long does it take to cover a distance of 1km?
Answers
Answer: 50 seconds
Explanation:
1km=1000m
Distance/speed=time
1000/20=50
50 seconds
Explanation:
Hey, there!!
Here,
speed (s)= 20m/s
distance (d)= 1km
= 1000m
now,
we have formula,
[tex]t = \frac{d}{s} [/tex]
putting value,
[tex]t = \frac{1000m}{20m/ s } [/tex]
cancelling the like terms,
[tex]t = 50s[/tex]
Therefore, the time is 50s.
Hope it helps...