A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 35880Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)
Answer:
Period is 86811.5 seconds.
Explanation:
[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]
[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]
A circuit consists of four 100W lamps
connected in parallel across a 230V supply.
Inadvertently, a voltmeter has been connected
in series with the lamps. The resistance of the
voltmeter is 15000 and that of the lamps
under the conditions stated is six times their
value when burning normally. What will be the
reading of the voltmeter?
Complete question is;
A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?
Answer:
150.42 V
Explanation:
We are told that the circuit consists of four 100W lamps.
We know that Power is given by the equation;
P = V²/R
Thus;
R = V²/P
Now, we are told that the four lamps are connected in parallel across a 230V supply.
Thus, V = 230 V
So resistance, R = 230²/100
R = 529 Ω
We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.
Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω
So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω
Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω
Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω
Thus;
circuit current = 230/2293.5 = 0.100283 A
Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V
CHECK THE COMPLETE QUESTION BELOW
A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?
Answer:
the reading of the voltmeter=150.4V
Explanation:
We can determine the wattage of a lamp using below expression:
: W = I² R....................eqn(1)
But fro ohms law V=IR
then I= V/R
If we substitute I into equation (1)
We have W= V²/R
But W= 100W
V= 230V
Then
W=220²/R
100 = 2302/R
R = 529 Ω
We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.
R = 6 × 529 = 3174 Ω
We can also calculate quivalent resistance of the abovefour lamps connected in parallel then
R = 3174/4
= 793.5 Ω
total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement
Then
total circuit resistance = 1500 + 793.5
= 2293.5 Ω
Then from ohms law again
I= V/R
circuit current = 230/2293.5 A
The reading of the voltage drop across the voltmeter
= 1500 × 230/2293.5
= 150.4V
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
RADIOWAVES of constant amplitude can be generated with
1) FET
2) filter
3)rectifier
4)oscillator
Answer:
4) oscillator
Explanation:
4) oscillator
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
There are two cells, one with OER as 2.5 and other as 7. Which cell is more sensitive to radiation?
1)The cell with OER 2.5
2)The Cell with OER 7
3)Both the cells
4)Insufficient data
Answer:
2)The Cell with OER 7
Explanation:
OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.
The generator in a purely inductive AC circuit has an angular frequency of 363 rad/s. If the maximum voltage is 169 V and the inductance is 0.0937 H, what is the rms current in the circuit
Answer:
The rms current in the circuit is 3.513 A
Explanation:
Given;
angular frequency of the inductor, ω = 363 rad/s
maximum voltage of the inductive AC, V₀ = 169 V
Inductance of the inductor, L = 0.0937 H
Inductive reactance is given by;
[tex]X_L = 2\pi f L= \omega L[/tex]
[tex]X_L = 363 *0.0937\\\\X_L = 34.0131 \ ohms[/tex]
The rms voltage is given by;
[tex]V_{rms} = \frac{V_o}{\sqrt{2} } \\\\V_{rms} =\frac{169}{\sqrt{2} } \\\\V_{rms} = 119.5 \ V[/tex]
The rms current in the circuit is given by;
[tex]I_{rms} = \frac{V_{rms}}{X_L} \\\\I_{rms} = \frac{119.5}{34.0131} \\\\I_{rms} = 3.513 \ A[/tex]
Therefore, the rms current in the circuit is 3.513 A
What is the difference between matter and energy
Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Do an Internet search to determine what minerals are extracted from the ground in order to manufacture the following products:
a. Stainless steel utensils
b. Cat litter
c. Tums brand antacid tablets
d. Lithium batteries
e. Aluminum beverage cans
Answer:
Raw materials are most times gotten from the earth through various forms of extraction procedures.
A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.
B) Cat litter comprises of ceramic products which is made up of clay.
C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.
D)Lithium batteries are made up of elements in the earth such as lithium and carbon.
E)Aluminum beverage cans are made up of aluminum extracted from the ground.
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]
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
0.99Hz
Explanation:
Using F= -mx ( spring force)
At equilibrium the gravitational force will be balanced by the spring force so mg= kx
K= mg/ 0.25 N/m
But
Frequency f= 1/2pi √g/0.25
Frequency is 0.99Hz
The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz
Frequency of oscillation based problem:What information do we have?
Length starched = 2.5 cm
F = Kx
We know that
F = mg
So,
mg = Kx
K/m = g/x
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]
Frequency of oscillation = 3.15 Hz
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In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interval of 0.001s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:
Answer:
The initial kinetic energy of the bullet is closest to 491.87 J
Explanation:
Given;
mass of bullet, m₁ = 18g = 0.018kg
mass of block, m₂ = 10kg
height moved by the block, h = 9 mm = 0.009 m
time taken for the bullet to travel through the block, t = 0.001s
let the initial velocity of the bullet = v₁
let the final velocity of the bullet = v₂
Apply the principle of conservation of linear momentum;
initial momentum = final momentum
0.018v₁ = v₂(0.018 + 10)
0.018v₁ = 10.018v₂ -----equation (1)
Apply the law of conservation of energy when the bullet lifts the block through 9mm
mgh = ¹/₂mv₂²
gh = ¹/₂v₂²
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.009)
v₂ = 0.42 m/s
Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;
0.018v₁ = 10.018v₂
0.018v₁ = 10.018(0.42)
0.018v₁ = 4.208
v₁ = 4.208 / 0.018
v₁ = 233.78 m/s
Now, determine the initial kinetic energy of the bullet;
K.E₁ = ¹/₂m₁v₁²
K.E₁ = ¹/₂(0.018)(233.78)²
K.E₁ = 491.87 J
Therefore, the initial kinetic energy of the bullet is closest to 491.87 J
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14 mm , the eyepiece a focal length of 21 mm .
A) Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer using two significant figures.
B) How tall is the image mentioned in part A? Express your answer using two significant figures.
C) If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens? Express your answer using two significant figures.
D) Under the conditions of part C, find the overall magnification of the microscope. Express your answer using two significant figures.
Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula
[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]
v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
What is objective lens?The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
By using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Putting the values
[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]
v = 210 mm .
B ) Magnification is the ratio of the size of the image to the size of the an object.
[tex]\rm magnification = \dfrac{v} { u}[/tex]
[tex]M= \dfrac{210} { 15}[/tex]
M= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
[tex]= 14 \times \dfrac{ 250} { 21}[/tex]
= 166.67
= 170 ( in two significant figures )
Hence all the answers are:
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
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b) Calculate the equivalent capacitance of the network shown below between the points A nd 'B', Given: C1 = C2 = 12uF.C3 = 7uF, CA = C5 = C6 =151F C6 =15uF
As No diagram attached I am taking all are connected in series
We know
[tex]\boxed{\sf C_{eq}=C_1+C_2\dots}[/tex]
[tex]\\ \sf \longmapsto C_{eq}=12+12+7+15+15+15[/tex]
[tex]\\ \sf \longmapsto C_{eq}=24+7+45[/tex]
[tex]\\ \sf \longmapsto C_{eq}=76\mu F[/tex]
a body accelerate uniformly from rest at the rate of 3meters per seconds for 8 sec . calculate the distance covered by the body during the acceleration
SOL
Answer:
96 m
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²
Δx = 96 m
radio waves are electromagnetic waves that travel at the speed of light 300 000 kilometers per second what is the wave length of FM radio waves received 100 megahertz on your radio dial
The wavelength of 100-MHz radio waves is 3 m, yet using the sensitivity of the resonant frequency to the magnetic field strength, details smaller than a millimeter can be imaged.Hope this helps you ❤️MaRk mE aS braiNliest ❤️
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards
Answer:
His angular velocity will increase.
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = [tex]I[/tex]'ω'
where
[tex]I[/tex]' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = [tex]mr'^{2}[/tex]
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].
From
[tex]I[/tex]'ω' = [tex]I[/tex]ω
since [tex]I[/tex] is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
Why are scientific models important?
Answer:
Scientific models are representations of objects, systems or events and are used as tools for understanding the natural world. Models use familiar objects to represent unfamiliar things. Models can help scientists communicate their ideas, understand processes, and make predictions.
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun."
Answer:
Net force = 2.3686 × 10^(20) N
Explanation:
To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.
Now, from standards;
Mass of earth;M_e = 5.98 × 10^(24) kg
Mass of moon;M_m = 7.36 × 10^(22) kg
Mass of sun;M_s = 1.99 × 10^(30) kg
Distance between the sun and earth;d_se = 1.5 × 10^(11) m
Distance between moon and earth;d_em = 3.84 × 10^(8) m
Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m
Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²
Now formula for gravitational force between the earth and the moon is;
F_em = (G × M_e × M_m)/(d_em)²
Plugging in relevant values, we have;
F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²
F_em = 1.9909 × 10^(20) N
Similarly, formula for gravitational force between the sun and moon is;
F_sm = (G × M_s × M_m)/(d_sm)²
Plugging in relevant values, we have;
F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×
7.36 × 10^(22))/(1496.96 × 10^(8))²
F_se = 4.3595 × 10^(20) N
Thus, net force = F_se - F_em
Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N
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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A man using a 70kg garden roller on a level surface, exerts a force of 200N at 45 degrees to the ground. find the vertical force of the roller on the ground if,
i.he pulls
ii.he pushes the roller
Answer:
i) 545.2 N upwards
ii) 828.2 N downwards
Explanation:
mass of the roller = 70 kg
force exerted = 200 N
angle the force makes with the ground ∅ = 45°
weight of the roller W = mg
where
m is the mass of the roller
g is the acceleration due to gravity = 9.81 m/s^2
weight of the roller = 70 x 9.81 = 686.7 N
The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°
==> F = 200 x 0.707 = 141.5 N
i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards
Resultant vertical force = 686.7 N - 141.5 N = 545.2 N upwards
ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards
Resultant vertical force = 686.7 N + 141.5 N = 828.2 N downwards
In the diagram, the amplitude of the wave is shown by:
A
B
C
D
Answer:
A.
Explanation:
Amplitude measures how much a wave rises or falls. This is illustrated by A.
In the diagram, the amplitude of the wave is shown by A.
What is Amplitude?The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value.There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.
The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.
Amplitude is a property that is unique to waves and oscillations.
Therefore, in the diagram, the amplitude of a wave is shown by A.
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An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?
Answer:
3.9E-8
Explanation:
We know that
Mv²/r = Bqv
So
r= mv/Bq
But E is 1/2mv² which is 5.53eV
m²v² =2m x 5.53eV
mv = √( 2 x 9.1E-31)
So
r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19
= 3.9x10-8cm
Explanation:
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?
Answer:
200cm
Explanation:
Answer:
100cm
Explanation:
Using
F= ( N/2L)(√T/u)
F1 will now be (0.5*2)( √600/0.015)
=> L( wavelength)= 200/2cm = 100cm
The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above
Answer:
b. Third Law entropy
Explanation:
Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature" tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.
In the question above, the correct answer is option b.