Answer:
Option A is the only valid statement.
Explanation:
A)The electric field intensity is defined by the relationship:
E= -ΔV/Δr.
Now, according to the relationship above, the electric field would be the negative gradient of electric potential. Now, if the electric potential is constant throughout the given region of space, then the change in electric potential would be ΔV=0.
Thu,E= 0.
So the answer is that, E will be zero in this case.
So, the statement is valid.
B) Statement not valid because the field is the gradient of the potential. Hence, the field would be zero in any region where the potential is constant. However, constant does not necessarily mean a value of zero. With that being said, we can always change the definition of the potential function by adding a constant, to thus make it zero there. But then the potential will no longer be zero at infinity or in any different “flat” regions.
C) Statement not valid because, for the fact that electric field is zero at a particular point, it doesn't necessarily
imply that the electric potential is zero at that point. A good example would be the case of two identical charges which are separated by some distance. At the midpoint between the charges, the
electric field due to the charges would be zero. However, the electric potential due to the charges at that same point would not be zero. Thus, the potential will either have two positive contributions, if the charges are positive, or two negative contributions, if the charges are negative.
(D) Statement is not valid because, for the fact that electric potential is zero at a particular point, it does not necessarily imply that the electric field is zero at that point. A good example would be the case of a dipole, which
has two charges of the same magnitude, but opposite sign, and are separated by some distance. At
the midpoint between the charges, the electric potential due to the charges would be zero, but the electric field due to the charges at that same point would not be zero.
A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released from the equilibrium position with an upward velocity of 6 meters per second. Find an equation of the motion.
Answer:
x(t) = -3sin2t
Explanation:
Given that
Spring force of, W = 720 N
Extension of the spring, s = 4 m
Attached mass to the spring, m = 45 kg
Velocity of, v = 6 m/s
The proper calculation is attached via the image below.
Final solution is x(t) = -3.sin2t
11. A seesaw sits in static equilibrium. A child with a mass of 30 kg sits 1 m away from a pivot point. Another child sits 0.75 m away from the pivot point on the opposite side. The second child's mass is _____ kg.
Answer:
40 kgExplanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
How fast must a 2500-kg elephant move to have the same kinetic energy as a 67.0-kg sprinter running at 15.0 m/s
Answer:
2.45 m/s
Explanation:
kinetic energy = 1/2 * m * v^2
then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2
by solving for x, X = 2.45 m/s
What will be the volume and density of stone if mass of stone is 10 gram .please tell the answer fast it's very urgent I will mark as a brain me answer if you will answer it correct.
Answer:
[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]
Explanation:
Assume the stone consists of basalt, which has a density of 3.0 g/cm³.
[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]
The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.
Answer and Explanation:
There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e. [tex]B \alpha \frac{1}{r^2}[/tex]
The magnetic field is a volume of vectors
And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit
Therefore for a closed path, we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.
Newton’s first law says that if motion changes, then a force is exerted. Describe a collision in terms of the forces exerted on both objects.
Answer:
In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Explanation:
In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.
Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.
Flower bed is filled with five types of flowers. Which placement of the flowers represents the highest entropy?
Answer:
B
Explanation:
ANSEWER :B IN ROWS ONLY
A 2.3kg bicycle wheel has a diameter of 50cm. What torque must you apply to take the wheel from 0rpm to 120rpm in 5.5s?
Answer:
τ = 0.26 N.m
Explanation:
First we find the moment of inertia of the wheel, by using the following formula:
I= mr²
where,
I = Moment of Inertia = ?
m = mass of wheel = 2.3 kg
r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m
Therefore,
I = (2.3 kg)(0.25 m)²
I = 0.115 kg.m²
Now, we find the angular acceleration of the wheel:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s
ωi = Initial Angular Velocity = 0 rad/s
t = time = 5.5 s
Therefore,
α = (12.56 rad/s - 0 rad/s)/(5.5 s)
α = 2.28 rad/s²
Now, the torque is given as:
Torque = τ = Iα
τ = (0.115 kg.m²)(2.28 rad/s²)
τ = 0.26 N.m
2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the
Answer:
for the first interference m = 1 y = 2,839 10-3 m
for the second interference m = 2 y = 5,678 10-3 m
Explanation:
The double slit interference phenomenon, for constructive interference is described by the expression
d sin θ = m λ
where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.
In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles
tan θ = y / L
with the angle it is small,
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
the distance between the central maximum and an interference line is
y = m λ L / d
let's reduce the magnitudes to the SI system
λ = 546 nm = 546 10⁻⁹ m
d = 0.25 mm = 0.25 10⁻³ m
let's substitute the values
y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³
y = m 2,839 10⁻³
the explicit value for a line depends on the value of the integer m, for example
for the first interference m = 1
the distance from the central maximum to the first line is y = 2,839 10-3 m
for the second interference m = 2
the distance from the central maximum to the second line is y = 5,678 10-3 m
If, the limits of the visible spectrum are approximately 3000 A.U. and 5000 A.U. respectively. Determine the angular breadth of the first order visible spectrum produced by a plane diffraction grating having 12000 lines per inch when light is incident normally on the grating.
Answer:
θ₁ = 0.04º , θ₂ = 0.00118º
Explanation:
The equation that describes the diffraction pattern of a network is
d sin θ = m λ
where the diffraction order is, in this case they indicate that the order
m = 1
θ = sin⁻¹ (λ / d)
Trfuvsmod ls inrsd fr ll red s SI units
d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm
the distance between two lines we can look for it with a direct proportions rule
If there are 4724 lines in a centimeter, the distance for two hundred is
d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm
let's calculate the angles
λ = 300 10-9 m
θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)
θ₁ = sin⁻¹ (7.08 10-4)
θ₁ = 0.04º
λ = 5000
θ₂ = sin-1 (500 10-9 / 4,2337 10-4)
θ₂ = 0.00118º
can I get help please?
Consider a race between the following three objects: object 1, a disk; object 2, a solid sphere; and object 3, a hollow spherical shell. All objects have the same mass and radius.
Required:
a. Rank the three objects in the order in which they finish the race. To rank objects that tie, overlap them.
b. Rank the objects in order of increasing kinetic energy at the bottom of the ramp. Rank objects from largest to smallest. To rank items as equivalent, overlap them.
Answer:
Since the angular acceleration of the objects will be proportional to the torque (due to gravity) acting on them and they will all experience the same torque their accelerations will be inversely proportional to their moments of inertia:
I disk = 1/2 M R^2
I sphere = 2/5 M R^2
I shell = 2/3 M R^2
Thus the sphere will experience the greatest angular acceleration and reach the bottom first, and then be followed by the disk and the shell.
By conservation of energy they will all have the same kinetic energy when they reach the bottom of the ramp.
(a) The ranking of the objects in order of how they will finish the race is
solid sphere > disk > hollow spherical shell
(b) The ranking of the objects in order of kinetic energy is
solid sphere > disk > hollow spherical shell
The moment of inertia of each object is calculated as follows;
disk: [tex]I = \frac{1} {2} MR^2[/tex]solid sphere: I = [tex]\frac{2}{5} MR^2[/tex]hollow spherical shell: I = [tex]\frac{2}{3} MR^2[/tex]The angular momentum of the objects is calculated as follows;
[tex]L =I \omega \\\\\omega = \frac{L}{I}[/tex]
The object with the least moment of inertia is will have the highest speed.
The ranking of the objects in order of how they will finish the race;
solid sphere > disk > hollow
The kinetic energy of the objects is calculated as follows;
[tex]K.E = \frac{1}{2} I \omega ^2[/tex]
The ranking of the objects in order of kinetic energy;
solid sphere > disk > hollow
Learn more here:https://brainly.com/question/15076457
There are two nearby point sources, each emitting a light wave of frequency f. When the frequency f is increased, how will the distance between troughs of constructive interference change?
Answer:
An increase in frequency will cause an increase in the number of lines per centimeter and a smaller distance between each consecutive line. That is the distance between each trough
Explanation:
This is because higher frequency light source should produce an interference pattern with more lines per centimeter in the pattern and a smaller spacing between lines.
Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.
Answer:
The wavelength is [tex]\lambda = 6.28 *10^{-7}=628 nm[/tex]
The frequency is [tex]f = 4.78 Hz[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 0.048 \ mm = 4.8 *0^{-5}\ m[/tex]
The distance from the screen is [tex]D = 6.50 \ m[/tex]
The distance between fringes is [tex]Y = 8.5 \ cm = 0.085 \ m[/tex]
Generally the distance between the fringes for a two slit interference is mathematically represented as
[tex]Y = \frac{\lambda * D}{d}[/tex]
=> [tex]\lambda = \frac{Y * d }{D}[/tex]
substituting values
[tex]\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }[/tex]
[tex]\lambda = 6.27 *10^{-7}=628 nm[/tex]
Generally the frequency of the light is mathematically represented as
[tex]f = \frac{c}{\lambda }[/tex]
where c is the speed of light with values
[tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]f = \frac{3.0*10^8}{6.28 *10^{-7}}[/tex]
[tex]f = 4.78 Hz[/tex]
A trolley going down an inclined plane has an acceleration of 2cm/s^2 What will be its velocity
3s after the start.
Answer:
[tex]V_{f}[/tex] = 6 cm/s
Explanation:
Given:
Acceleration = a = 2 cm/s²
Time = t = 3s
Initial Velocity = [tex]V_{i}[/tex] = 0 cm/s
Required:
Velocity = [tex]V_{f}[/tex] = ?
Formula:
a = [tex]\frac{V_{f}-V_{i}}{t}[/tex]
Solution:
2 = [tex]\frac{V_{f}-0}{3}[/tex]
=> [tex]V_{f}[/tex] = 2*3
=> [tex]V_{f}[/tex] = 6 cm/s
A tennis player tosses a tennis ball straight up and then catches it after 1.91 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction ---Select--- (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m
Explanation:
(a) The acceleration is 9.8 m/s² downwards.
(b) At the maximum height, the velocity is 0 m/s.
(c) v = at + v₀
0 m/s = (-9.8 m/s²) (1.91 s) + v₀
v₀ = 18.7 m/s
(d) Δy = vt − ½ at²
Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²
Δy = 17.9 m
If, instead, the ball is revolved so that its speed is 3.7 m/s, what angle does the cord make with the vertical?
Complete Question:
A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.
(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its
speed is 3.7 m/s, what angle does the cord make with the vertical?
(Check attached image for the diagram.)
Answer:
(a) The ball’s speed, v = 2.06 m/s
(b) The angle the cord makes with the vertical is 50.40⁰
Explanation:
If the ball is revolved in a horizontal plane, it will form a circular trajectory,
the radius of the circle, R = Lsinθ
where;
L is length of the string
The force acting on the ball is given as;
F = mgtanθ
This above is also equal to centripetal force;
[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]
(b) when the speed is 3.7 m/s
[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]
[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]
Therefore, the angle the cord makes with the vertical is 50.40⁰
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct
Complete question:
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A. the interior field points in a direction parallel to the exterior field
B. There is no electric field on the interior of the conducting sphere.
C. The interior field points in a direction perpendicular to the exterior field.
D. the interior field points in a direction opposite to the exterior field.
Answer:
B. There is no electric field on the interior of the conducting sphere.
Explanation:
Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).
Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of free charges in the spherical conductor in response to any field until its neutralization.
Option B is correct.
There is no electric field on the interior of the conducting sphere.
Describe the relationship between the density of electric field lines and the strength of the electric field?
Answer:
The greater the density of the electric field lines the stronger the electric field and vice versa
Explanation:
Electric field can be defined as the region where an electric force is experienced by a charged body. A charged body experiences a force whenever it is positioned close to another charged body.
An electric field may be described in terms of lines of force which represent the direction of a small positive charge placed at that point assuming that the charge is so small that it does not change appreciably in the presence of another charge. Arrows on the lines of force indicate the direction of the electric field.
The lines of force are indicated in such a way that the strength of the electric field is shown by the number or density of electric field lines crossing a unit area perpendicular to the lines. Hence, the greater the density of the electric field lines the stronger the the electric field and vice versa
At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 36.0 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?
Answer:
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.
Explanation:
The energy stored in the inductor is given as
E₁ = ½LI²
The rate at which energy is stored in the inductor is
(dE₁/dt) = (d/dt) (½LI²)
Since L is a constant
(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)
(dE₁/dt) = LI (dI/dt)
Rate of Energy dissipated in a resistor = Power = I²R
(dE₂/dt) = I²R
When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field
(dE₁/dt) = (dE₂/dt)
OK (dI/dt) = I²R
L (dI/dt) = IR
Current in a this kind of series setup of inductor and resistor at any time, t, is given as
I = (V/R) (1 - e⁻ᵏᵗ)
k = (1/time constant) = (R/L)
(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ
L (dI/dt) = IR
L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)
V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)
e⁻ᵏᵗ = 1 - e⁻ᵏᵗ
2 e⁻ᵏᵗ = 1
e⁻ᵏᵗ = (1/2) = 0.5
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = -0.69315
- kt = -0.69315
kt = 0.69315
k = (1/time constant)
Time constant = 36.0 ms = 0.036 s
k = (1/0.036) = 27.78
27.78t = 0.69315
t = (0.69315/27.78) = 0.02495 = 24.95 ms
Hope this Helps!!!
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Answer:
[tex]v_{2}=3.5 m/s[/tex]
Explanation:
Using the conservation of energy we have:
[tex]\frac{1}{2}mv^{2}=mgh[/tex]
Let's solve it for v:
[tex]v=\sqrt{2gh}[/tex]
So the speed at the lowest point is [tex]v=7 m/s[/tex]
Now, using the conservation of momentum we have:
[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]
[tex]v_{2}=\frac{1*7}{2}[/tex]
Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]
I hope it helps you!
Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the earth observer, while ship B moves at 0.60c relative to the same observer. What speed does the captain of ship A report for the speed of ship B
Answer:
0.80 c
Explanation:
The computation of speed is shown below:-
Here, The speed of the captain ship A report for speed of the ship B which is
[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]
where
[tex]S_A[/tex] indicates the speed of the ship A
[tex]S_B[/tex] indicates the speed of the ship B
and
C indicates the velocity of life
now we will Substitute 0.40c for A and 0.60 for B in the equation which is
[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]
after solving the above equation we will get
0.80 c
So, The correct answer is 0.80c
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement does the puck have to travel in order to make it to the net?
Answer:
x=22.57 m
Explanation:
Given that
35 m in W of S
angle = 40 degrees
25 m in east
From the diagram
The angle
[tex]\theta=90-40=50^o[/tex]
From the triangle OAB
[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]
[tex]1340.57=35^2+25^2-x^2[/tex]
x=22.57 m
Therefore the answer of the above problem will be 22.57 m
A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed
Answer:
The new angular speed is [tex]w = 6 \ rev/s[/tex]
Explanation:
From the question we are told that
The angular velocity of the spin is [tex]w_o = 3 \ rev/s[/tex]
The original moment of inertia is [tex]I_o[/tex]
The new moment of inertia is [tex]I =\frac{I_o}{2}[/tex]
Generally angular momentum is mathematically represented as
[tex]L = I * w[/tex]
Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so
[tex]I * w = constant[/tex]
=> [tex]I_o * w _o = I * w[/tex]
where w is the new angular speed
So
[tex]I_o * 3 = \frac{I_o}{2} * w[/tex]
=> [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]
=> [tex]w = 6 \ rev/s[/tex]
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 10^2 N/C and separation between the charged plates is 2.0 cm. a.) Determine the horizontal distance traveled by the electron when it hits the plate. b.)Determine the velocity of the electron as it strikes the plate.
Answer:
Explanation:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4*10^6[2*2*10^-2/7*10^14]^1/2
=3*10^-2=3 cm
now we find the velocity f the electron strike the plate
v^2-(4*10^6)^2=2*7*10^14*2*10^-2
v^2=16*10^12+28*10^12
v^2=44*10^12
speed after hits =>V=6.6*10^6 m/s
Muons are elementary particles that are formed high in the atmosphere by the interactions of cosmic rays with atomic nuclei. Muons are radioactive and have average lifetimes of about two-millionths of a second. Even though they travel at almost the speed of light, they have so far to travel through the atmosphere that very few should be detected at sea level - at least according to classical physics. Laboratory measurements, however, show that muons in great number do reach the earth's surface. What is the explanation?
Answer:
Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.
Explanation:
Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.
what is the preferred method of using percentage data by using a circle divided into sections
Answer:
A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole
Explanation:
pie charts are a useful way to organize data in order to see the size of components relative to the whole.
Point X is midway between the charges. In what section of the line will there be a point where the resultant electric field is zero?
Answer:
I believe the answer is in fact section (VW) on the line where the electric field result will be zero.
Explanation:
The direction of the electric field due to a positive charge is away from it and the direction of the electric field due to a negative one is towards it.
A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the mass is released from rest at x = 10.0 cm. ( That is, the spring is stretched by 10.0 cm.) (a) Determine the frequency of the oscillations. (b) Determine the maximum speed of the mass. Where dos the maximum speed occur? (c) Determine the maximum acceleration of the mass. Where does the maximum acceleration occur? (d) Determine the total energy of teh oscillating system. (e) Express the displacement as a function of time.
Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by:
[tex]v_{max}=\omega A=2\pi f A[/tex] (2)
A: amplitude of the oscillations = 10.0cm = 0.10m
[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by:
[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]
[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is:
[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]
The total energy is 0.1J
(e) The displacement as a function of time is:
[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]