Answer:
The right solution will be the "2v".
Explanation:
For something like an object underneath pure rolling the speed at any point is calculated by:
⇒ [tex]v_{rolling}=v_{translational}+v_{rotational}[/tex]
Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:
⇒ [tex]\omega=\frac{v_{translational}}{r}[/tex]
Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:
⇒ [tex]v_{rotational}=\omega \times r[/tex]
On substituting the estimated values, we get
⇒ [tex]=\frac{v_{translational}}{r} \times r[/tex]
⇒ [tex]=v_{translational}[/tex]
So that the velocity will be:
⇒ [tex]v_{rolling}=v+v[/tex]
⇒ [tex]=2v[/tex]
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!!!
The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
(a) If it takes 3.00×10-13 s for a nitrogen molecule to hit a wall and rebound with the same speed but
moving in the opposite direction, what is the average acceleration of the molecule during this time
interval?
(b) What average force does the molecule exert on the wall?
(c) What is the total force exerted
on the wall if it is struck by 1015 such molecules all at once?
Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m [tex]v_{f}[/tex] - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
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
can I get help please?
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
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
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.
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
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
If an object on Earth weighs 100N what is its weight in pounds?
Answer:
10.2 lbs
Explanation:
m=F/g
m=100N/9.8
m=10.2040816 lbs
The weight of the object in pounds will be 220.46 pounds m/s².
We have an object on Earth whose weight is 100 N.
We have to determine its weight in pounds.
Pounds is used to represent which Physical quantity ?Pounds is used to represent the mass of the body.
According tot the question -
Weight on earth = 100 N = 100 Kg . m/s²
1 Kg = 2.2046 Pounds
Therefore, the weight on earth in pounds will be = 100 x 2.2046 pounds m/s² = 220.46 pounds m/s².
Hence, the weight of the object in pounds will be 220.46 pounds m/s².
To solve more questions on Unit conversion, visit the link below-
https://brainly.com/question/14024722
#SPJ2
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.
In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 × 10 6 m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit is 0.020 m. What is the mass of the particle?
Answer:
The mass is [tex]m =6.4*10^{-28} \ kg[/tex]
Explanation:
From the question we are told that
The speed of the charge is [tex]v = 1.0 *10^{6} \ m/s[/tex]
The magnetic field is [tex]B = 0.20 \ T[/tex]
The radius is [tex]r = 0.02 \ m[/tex]
The value of the charge is [tex]e = 1.60 *10^{-19} \ C[/tex]
The centripetal acting on the charge moving in the circular orbit is mathematically represented as
[tex]F_c = \frac{mv^2}{r }[/tex]
Now this centripetal force is due to the force exerted on the charge by the magnetic field on the charge which is mathematically represented as
[tex]F_m = qv B sin\theta[/tex]
At the maximum of this magnetic force [tex]\theta = 90 ^o[/tex]
So
[tex]F_m = e v B sin(90)[/tex]
[tex]F_m = e v B[/tex]
Now given that it is this magnetic force that is causing the circular motion we have that
[tex]F_c = F_m[/tex]
=> [tex]\frac{mv^2}{r } = ev B[/tex]
=> [tex]m = \frac{e * B * r }{v }[/tex]
substituting values
[tex]m = \frac{ 1.60 *10^{-19} * 0.20 * 0.020 }{1.0*10^{6} }[/tex]
[tex]m =6.4*10^{-28} \ kg[/tex]
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
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.
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
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
Explanation:
Power of electrical circuit = I² R where I is current and R is resistance
Putting the given data
1.64 x 10⁵ = 12² x R
R = 1.139 x 10³ ohm
= 1139 ohm .
Answer:
r=1140ohms
Explanation: plato family
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.
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.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
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
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 skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has an inclination of 35 °
a) At what point does the skier make contact again with the ground?
Answer:
107 m down the incline
Explanation:
Given:
v₀ₓ = 25 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
-Δy/Δx = tan 35°
Find: d
First, find Δy and Δx in terms of t.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) t + ½ (-10 m/s²) t²
Δy = -5t²
Δx = v₀ₓ t + ½ aₓ t²
Δx = (25 m/s) t + ½ (0 m/s²) t²
Δx = 25t
Substitute:
-(-5t²) / (25t) = tan 35°
t/5 = tan 35°
t = 5 tan 35°
t ≈ 3.50 s
Now find Δy and Δx.
Δy ≈ -61.3 m
Δx ≈ 87.5 m
Therefore, the distance down the incline is:
d = √(x² + y²)
d ≈ 107 m
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 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!
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
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
What is the relation between the direction of electric field and equipotential line at the same point
Answer:
The direction of electric field and equipotential line at the same point are always PERPENDICULAR TO THE ELECTRIC FIELD.
Explanation:
Equipotential surface is a three dimensional part of equipotential lines.
Equipotential lines are a type of contour lines that is use to trace lines that have the same altitude on the map and the altitude is the electric potential.
Equipotential lines are always perpendicular to electric potential because the lines creates three dimension equipotential surface.
The direction of the electric field and equipotential line at the same point is always perpendicular.
Equipotential lines:The electric field and the electric potential both are dependent on distance. As we move farther from the source generating the electric field, the electric field strength, as well as the electric potential strength, decreases.
The electric field is inversely proportional to the square of the distance from the source, while the electric potential is inversely proportional to the distance itself.
The equipotential surface is a surface on which every point is at the same distance from the source so that each point is at the same potential.
Now, we can make an equipotential line joining these points. Since it is fixed at a surface, we can not move farther or closer to the source, because there will be a change in the distance and the potential will change.
So these lines are always perpendicular to the lines representing the electric field, which travel towards or away from the source
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Suppose you have a box of identical resistors. Each resistor in the box has a resistance of 20 Ohms. Explain how to connect them together so the total resistance of the network is 25 Ohms. Use the minimum number of resistors in your design.
Answer:
By using 4 resistors in parallel and followed by a resistor in series.
Explanation:
Let consider 5 resistors with a resistance of 20 ohms each, the total resistance is equal to 25 ohms when the first four resistance are settled in parallel and is followed by a resistor is series. The calculations are presented below:
1st Stage (4 resistors in parallel)
[tex]R_{eq,1} = \frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2} }+\frac{1}{R_{3}}+\frac{1}{R_{4}}}[/tex]
If [tex]R_{1} = R_{2} = R_{3} = R_{4} = 20\,\Omega[/tex], then:
[tex]R_{eq,1} = \frac{1}{\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega}+\frac{1}{20\,\Omega} }[/tex]
[tex]R_{eq,1} = \frac{1}{\frac{4}{20\,\Omega} }[/tex]
[tex]R_{eq,1} = 5\,\Omega[/tex]
2nd Stage (1 resistor in series)
[tex]R_{5} = 20\,\Omega[/tex]
Now, the equivalent resistance is:
[tex]R_{eq,2} = R_{eq,1} + R_{5}[/tex]
[tex]R_{eq,2} = 5\,\Omega + 20\,\Omega[/tex]
[tex]R_{eq,2} = 25\,\Omega[/tex]
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
A small motor is mounted on the axis of a space probe with its rotor (the rotating part of the motor) parallel to the axis of the probe. Its function is to control the rotational position of the probe about the axis. The moment of inertia of the probe is [07] times that of the rotor. Initially, the probe and rotor are at rest. The motor is turned on and after some period of time, the probe is seen to have rotated by positive 32.6 degrees. Through how many revolutions has the rotor turned
Answer:
0.634 rev
Explanation:
It is stated that,
The moment of inertia of the probe = 7 times the moment of inertia of the rotor
After running for some time, the probe is seen to have rotated through positive 32.6°.
According to the laws of conservation of angular momentum, the angular momentum of the probe must be equal to the angular momentum of the rotor.
==> Irωr = Ipωp ....... equ 1
integrating the angular speed ω with respect to time t leaves us with
Ir∅r = Ip∅p ...... equ 2
where,
Ir = moment of inertia of the rotor
ωr = angular speed of the rotor
Ip = moment of inertia of the rotor
ωp = angular speed of the probe
∅r = angular position of the rotor
∅p = angular position of the probe
equ 2 can be rewritten as
Ip/Ir = ∅r/∅p
from the statement, Ip/Ir = 7
therefore,
7 = ∅r/∅p = ∅r/32.6
∅r = 7 x 32,6 = 228.2°
converting to rev = [tex]\frac{228.2}{360 }[/tex] = 0.634 rev