Answer:
2005 J
Explanation:
500 cal × (4.18 J/cal) = 2090 J
ΔU = q + w
ΔU = 2090 J − 85 J
ΔU = 2005 J
A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.70 s. Find the force constant of the spring.
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m
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.
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.
Question:
A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)
Answer:
3.22 x 10⁻⁴ m/s
Explanation:
The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;
v = [tex]\frac{I}{nqA}[/tex]
Where;
n = number of free electrons per cubic meter
q = electron charge
A = cross-sectional area of the wire
First let's calculate the number of free electrons per cubic meter (n)
Known constants:
density of copper, ρ = 8.95 x 10³kg/m³
molar mass of copper, M = 63.5 x 10⁻³kg/mol
Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol
But;
The number of copper atoms, N, per cubic meter is given by;
N = (Nₐ x ρ / M) -------------(ii)
Substitute the values of Nₐ, ρ and M into equation (ii) as follows;
N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³
N = 8.49 x 10²⁸ atom/m³
Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;
n = 8.49 x 10²⁸ electrons/m³
Now let's calculate the drift electron
Known values from question:
A = 5.261 mm² = 5.261 x 10⁻⁶m²
I = 23A
q = 1.6 x 10⁻¹⁹C
Substitute these values into equation (i) as follows;
v = [tex]\frac{I}{nqA}[/tex]
v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]
v = 3.22 x 10⁻⁴ m/s
Therefore, the drift electron is 3.22 x 10⁻⁴ m/s
A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about point A, that is, the point of contact between the front tire of the forklift and the ground
Answer:
The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]
Explanation:
mass of the crate = [tex]M_{C}[/tex]
speed of forklift = [tex]V_{1}[/tex]
The distance between the center of the mass and the point A = d
Recall that the angular moment is the moment of the momentum.
[tex]L = P*d[/tex] ..... equ 1
where L is the angular momentum,
P is the momentum of the system,
d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image
Also, we know that the momentum P is the product of mass and velocity
P = mv ....equ 2
in this case, the mass = [tex]M_{C}[/tex]
the velocity = [tex]V_{1}[/tex]
therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]
we substitute equation 2 into equation 1 to give
[tex]L = M_{C} V_{1} d[/tex]
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 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
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
A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting
Answer:
Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m
Explanation:
The maximum electric field in the beam = 2E
cross-sectional area of beam = A
The intensity of an electromagnetic wave with electric field is
I = cε[tex]E_{0} ^{2}[/tex]/2
for [tex]E_{0}[/tex] = 2E
I = 2cε[tex]E^{2}[/tex] ....equ 1
where
I is the intensity
c is the speed of light
ε is the permeability of free space
[tex]E_{0}[/tex] is electric field
Radiation pressure of an electromagnetic wave on an absorbing surface is given as
P = I/c
substituting for I from above equ 1. we have
P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex] ....equ 2
Also, pressure P = F/A
therefore,
F = PA ....equ 3
where
F is the force
P is pressure
A is cross-sectional area
substitute equ 2 into equ 3, we have
F = 2ε[tex]E^{2}[/tex]A
force on a body = mass x acceleration.
that is
F = ma
therefore,
a = F/m
acceleration of the asteroid will then be
a = 2ε[tex]E^{2}[/tex]A/m
where m is the mass of the asteroid.
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]
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.
Answer:
Velocity.
Explanation:
Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.
As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:
Horizontal range: As per expression:
R= ([tex]u^{2}[/tex]*sin2θ)/g
the range depending on the square of the initial velocity.
Maximum height: As per expression:
H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g
the maximum distance also depends upon square of the initial velocity.
can I get help please?
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
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.
What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level = [tex]1 \times 10-12 W/m^2[/tex]
Now
Intensity level ( or Loudness)is
[tex]L = log10 \frac{I}{10}[/tex]
[tex]L = log10 \frac{10}{1\times 10^{-12}}[/tex]
[tex]L = log10 \times 1013[/tex]
[tex]= 13 \times 1 ( log10(10) = 1)[/tex]
Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels
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!
An object of mass m is traveling in a circle with centripetal force f c if the velocity of the object is v what is the radius of the circle
Answer:
m=Fcr/v^2
Explanation:
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
The position of a helicopter of weight=2.75 X 10^5N is r=(0.020m/s3)(t3)i + (2.2 m/s)(t)j - (0.060m/s2)(t2)k. What is the net force on the helicopter at t=5.0s?
Answer:
F = 17170.16 N = 17.17 KN
Explanation:
First we need to find the mass of helicopter by using its weight:
Weight = mg
2.75 x 10⁵ N = m(9.8 m/s²)
m = (2.75 x 10⁵ N)/(9.8 m/s²)
m = 28061.22 kg
Now, we find acceleration. We have position vector as:
r = (0.02 m/s³)(t³)i + (2.2 m/s)(t)j - (0.06 m/s²)(t²)k
taking its derivative twice, we can find acceleration:
a = (3)(2)(0.02 m/s³)(t)i + (0)j - (2)(1)(0.06 m/s²)k
a = (0.12 m/s³)(t)i - (0.12 m/s²)k
at, t = 5 sec
a = (0.12 m/s³)(5 s)i - (0.12 m/s²)k
a = (0.6 m/s²) i - (0.12 m/s²) k
Now, the magnitude of acceleration will be:
a = √[(0.6)² + (-0.12)²]
a = 0.61 m/s²
So, from Newton's Second Law, the net force on helicopter is given as:
F = ma
F = (28061.22 kg)(0.61 m/s²)
F = 17170.16 N = 17.17 KN
Suppose the frequency of a note on an organ is 18 Hz. What is the shortest organ pipe with both ends open that will resonate at this frequency
Answer:
9.53 m
Explanation:
The computation of shortest organ pipe with both ends open that will resonate at this frequency is shown below:-
[tex]\lambda = \frac{velocity}{frequency}[/tex]
[tex]= \frac{343}{18}[/tex]
= 19.06 m
Now the
Shortest organ pipe with both ends open is
= [tex]\frac{\lambda}{2}[/tex]
[tex]= \frac{19.06}{2}[/tex]
= 9.53 m
Basically we applied the above formulas so that first we easily determined the shortest organ pipe for both ends at this frequency
What do energy advisors mean by the phrase "the greenest kilowatt is the one you never use?"
Answer:
"Energy deficiency, no coal-burning, no-cost mining pollution" is the correct answer.
Explanation:
“The greenest kilowatt-hour seems to be the one this really doesn't should use,” explained Joe Stepenovitch, co-owner as well as COO of something like the electricity IQ Group. Whether a kilowatt becomes generated is far less essential instead of not needing to do something with it.It, therefore, reduces operational costs, appeals to progressives and green-conscious consumers, prepares the business for impending emissions reductions policy caps, as well as coincides with you including an imminent future focused on renewable energy sources.What If? What would be the new angular momentum of the system (in kg · m2/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)
Complete question is;
(a) A light, rigid rod of length, l = 1.00 m joins two particles, with masses m = 4.00 kg and m, = 3.00 kg, at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 6.4 m/s. (Enter the magnitude to at least two decimal places in kg. m/s.)
(b) What If? What would be the new angular momentum of the system (in kg. m/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.)
Answer:
A) L = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis
B) L = 3.36 kg.m²/s
Explanation:
The image is missing, so i have attached it.
Formula for the moment of Inertia would be; I = mr²
m1 = 4 kg
m2 = 3 kg
r = 1/2 = 0.5 m
So, sum of moment of inertia for the 2 masses would be;
I = (4 × 0.5²) + (3 × 0.5²)
I = 1.75 kg.m²
Now, angular velocity is given by the formula;
ω = v/r
We are given v = 6.4 m/s
So;
ω = 6.4/0.5
ω = 12.8 rad/s
Now, let's find angular momentum.
Angular momentum; L = Iω
L = 1.75 × 12.8
L = 22.4 kg.m²/s
Now, using the right hand rule, the direction will be along the positive(+ve) z-axis.
B) Now, the new diameter is 15 cm = 0.15 m
Thus,
radius;r = 0.15/2 = 0.075 m
Similar to a above;
I = (4 × 0.075²) + (3 × 0.075²)
I = 0.039375 kg.m²
ω = v/r
We are given v = 6.4 m/s
ω = 6.4/0.075
ω = 85.33 rad/s
Angular momentum; L = Iω
L = 0.039375 × 85.33
L = 3.36 kg.m²/s
The Complete question is;
(a) A light, rigid rod of length, l = 1.00 m joins two particles, with masses m = 4.00 kg and m, = 3.00 kg, at its ends. The combination rotates in the xy-plane about a pivot through the center of the rod (see figure below). Determine the angular momentum of the system about the origin when the speed of each particle is 6.4 m/s. (Enter the magnitude to at least two decimal places in kg. m/s.)
(b) What If? What would be the new angular momentum of the system (in kg. m/s) if each of the masses were instead a solid sphere 15.0 cm in diameter? (Round your answer to at least two decimal places.) Image check below.
A) L is = 22.4 kg.m²/s and it's direction will be along the positive(+ve) z-axis
B) L is = 3.36 kg.m²/s
What is Angular momentum?
We are applying the Formula for the moment of Inertia would be; I = mr²
Then, m1 = 4 kg
After that, m2 is = 3 kg
Now, r = 1/2 = 0.5 m
So, When the sum of the moment of inertia for the 2 masses would be;
Then, I = (4 × 0.5²) + (3 × 0.5²)
After that, I = 1.75 kg.m²
Now, when the angular velocity is given by the formula;
Then, ω = v/r
We are given v is = 6.4 m/s
Then, ω = 6.4/0.5
After that, ω = 12.8 rad/s
Now, let's find the angular momentum.
Then, the Angular momentum; L = Iω
L is = 1.75 × 12.8
Theregore, L = 22.4 kg.m²/s
Now, we are using the right-hand rule, the direction will be along the positive(+ve) z-axis.
B) Now, wehn the new diameter is 15 cm = 0.15 m
Thus, radius;r = 0.15/2 = 0.075 m
Then, Similar to a above;
After that, I = (4 × 0.075²) + (3 × 0.075²)
Now, I = 0.039375 kg.m²
Then, ω = v/r
We are given v = 6.4 m/s
ω is = 6.4/0.075
ω is = 85.33 rad/s
When the Angular momentum; L = Iω
Then, L = 0.039375 × 85.33
Therefore, L = 3.36 kg.m²/s
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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
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
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A projectile is launched with a velocity of 13.2m/s at an angle of 37 degrees above the horizontal. What is the horizontal component of the projectiles velocity 1s after the object is fired. PLS HURRY THIS IS TIMEd
Answer:
horizontal component = 10.54m/s
Explanation:
horizontal component = 13.2cos37°
horizontal component = 10.54m/s
A platypus foraging for prey can detect an electric field as small as 0.002 N/C. Part A To give an idea of sensitivity of the platypus's electric sense, how far from a 50nC point charge does the field have this magnitude
Answer:
474.34 m
Explanation:
From the question,
E = kq/r²................. Equation 1
Where E = Electric Field, k = coulomb's constant, q = Charge, r = distance.
Make r the subject of the equation
r = √(kq/E)............ Equation 2
Given: q = 50 nC = 50×10⁻⁹ C, E = 0.002 N/C
Constant: k = 9×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(50×10⁻⁹×9×10⁹/0.002)
r = √(450/0.002)
r = √(225000)
r = 474.34 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
A skydiver falls toward the ground at a constant velocity. Which statement best applies Newton’s laws of motion to explain the skydiver’s motion?
Answer:
A: An upward force balances the downward force of gravity on the skydiver.
Explanation:
on edge! hope this helps!!~ (⌒▽⌒)☆
Two people push on a large gate as shown on the view from above in the diagram. If the moment of inertia of the gate is 90 kgm2, what is the resulting angular acceleration of the gate?
Answer:
1ft per second
Explanation:
Physics is on my side!!!!!!!!!!
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 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]