A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of the point charge and the total charge on shell were, respectively:
Complete question is;
A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:
+5q; 0
-6q; +2q
+2q; -3q
-4q; +12q
Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:
a. 2, 3, 1, 4
b. 1, 2, 3, 4
c. 2, 4, 3, 1
d. 1, 3, 4, 2
Answer:
c. 2, 4, 3, 1
Explanation:
In this question, we can say that;
q_in = q_b
Where;
q_in is the charge on the inner surface of the shell
q_b is the point charge on the shell.
Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.
Thus;
- For +5q; 0:
q_in = -(+5q)
q_in = -5q
- For -6q; +2q :
q_in = - (-6q)
q_in = +6q
- For +2q; -3q :
q_in = -(+2q)
q_in = -2q
- For -4q; +12q:
q_in = -(-4q)
q_in = +4q
Ranking the most positive to the least positive ones, we have;
+6q, +4q, -2q, -5q
This corresponds to options;
2, 4, 3, 1
A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device
Answer:
Toaster = I = 14.67 A
Frying Pan = 11.83 A
Lamp = 0.71 A
Explanation:
The electric power is given as:
[tex]P = VI\\\\I = \frac{P}{V}[/tex]
where,
I = current
P = Power
V = Voltage = 120 V
FOR TOASTER:
P = 1760 W
Therefore,
[tex]I = \frac{1760\ W}{120\ V}[/tex]
I = 14.67 A
FOR FRYING PAN:
P = 1420 W
Therefore,
[tex]I = \frac{1420\ W}{120\ V}[/tex]
I = 11.83 A
FOR LAMP:
P = 85 W
Therefore,
[tex]I = \frac{85\ W}{120\ V}[/tex]
I = 0.71 A
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
Differences between angle of twist and angle of shear
Answer:
idek
Explanation:
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.
Answer:
non biodegradable
Explanation:
It is non biodegradable because plastic cannot dispose off easily ..
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
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A random sample of 22 lunch orders at Noodles & Company showed a mean bill of $10.26
with a standard deviation of $5.21. Find the 99 percent confidence interval for the mean bill of
all lunch orders.
Answer:
(7.115 ; 13.405)
Explanation:
Given :
Sample size, n = 22
Mean bill, μ = 10.26
Standard deviation, s = 5.21
To obtain the 99% confidence interval for the mean bill of all orders ;
Mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 99%, df = n-1, 22 - 1 = 21
Tcritical = 2.831
Margin of Error = 2.831 * (5.21/√22) = 3.145
Confidence interval = 10.26 ± 3.145
Lower boundary = 10.26 - 3.145 = 7.115
Upper boundary = 10.26 + 3.145 = 13.405
Confidence interval :
(7.115 ; 13.405)
A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?
Answer:
The answer is "151.25 J and -547.64 J".
Explanation:
[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]
Using formula:
[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]
[tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]
[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]
Calculating the Work by net force
[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]
The above work is converted into thermal energy.
Now,
[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
Physical quantities expresed only by their magnitude is
Answer:
Scalar quantity is the Physical quantity expresed only by their magnitude.
What is the effect on range and maximum height of a projectile as the launch height, launch speed, and launch angle are increased?
Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
Write one advantage of MKS system over CGS system.
Name the electrolyte in the chemical method of generating electricity
why is unit of power is called derived unit?
Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.
A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the surface of the earth). What is the direction of the magnetic field?
South
Explanation:
The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by
[tex]\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}[/tex]
and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by
[tex]K = 0.5 mv^2[/tex]
(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
(a) The kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b) The final speed of both the puck A and B are same.
Let the mass of puck A is m and the mass of puck B is 2 m.
Initial speed for both the pucks is same as u and the distance is same for both is s.
Let the tension is T for same.
Then, the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
(a)
As the speed is same, so the kinetic energy depends on the mass.
Then,
[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]
So, kinetic energy of A : Kinetic energy of B = 1 : 2.
Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b)
The final speed for the puck is given as,
v = s/t
here, s is the distance covered.
Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.
Thus, we can conclude that the final speed of both the puck A and B are same.
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A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.
What is efficiency?An efficiency of a heat engine is the ratio of the work done and heat supplied.
Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.
The efficiency is written as,
η= W / Qs.
The work done is W= Qs - Qr, where Qr is the rejected heat.
The heat rejected can be represented as
Qr = W ( 1/η -1)
Substituting the value into the equation, we get the rejected heat.
Qr = 2510 (1/0.22 -1)
Qr = 8900 Joules.
Thus, the heat rejected by the engine is 8900 Joules.
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If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring
Answer:
the natural length of the spring is 9 cm
Explanation:
let the natural length of the spring = L
For each of the work done, we set up an integral equation;
[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]
The second equation of work done is set up as follows;
[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]
solve equation (1) and equation (2) together;
[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]
[tex]l = 9 \ cm[/tex]
Therefore, the natural length of the spring is 9 cm
Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.
Answer:
F/2
Explanation:
In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant
Hence;
F= KQ1Q2/r^2 ------(1)
Where the charge on Q1 is doubled and the distance separating the charges is also doubled;
F= K2Q1 Q2/(2r)^2
F2= 2KQ1Q2/4r^2 ----(2)
F2= F/2
Comparing (1) and (2)
The magnitude of force acting on each of the two particles is;
F= F/2
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
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