Answer:
5.1 rad/s
Explanation:
Mechanical energy of the system is conserved since no external work is done on the sphere.
[tex]mgh = mv^2/2 + I\omega^2/2[/tex]
Substituting v = ωr and I = 2 m r^2/5, we get,
=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]
=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]
=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]
=> [tex]gh = 7\omega^2 r^2/10[/tex]
=> [tex]\omega r = (10gh/7)^{1/2}[/tex]
=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s
A truck has a bed that is 4.50 metres long,and 2.50 metres wide, and 1.50 metres high. What is maximum volume of sand can the truck carry within this dimensions?
Answer:
since the bed is a cuboid, we find the volume by L×W×H
4.50 × 2.50 × 1.50 = 16.875m³
HOPE THIS HELPS
16)
Gamma
rays
X-rays
UV
Infrared
Micro-
waves
Radio
waves
Visible light
Light is an electromagnetic wave and it has a place on the electromagnetic spectrum based on it energy and
wavelength. How does light's energy compare to the energy of other forms of electromagnetic waves?
A)
Light is less energetic than X-rays.
B)
Light is more energetic than X-rays.
Light is the least energetic electromagnetic wave.
D)
Light is the most energetic electromagnetic wave.
Answer:
Light is less energetic than X-rays.
Explanation:
The electromagnetic spectrum refers to the range of wavelengths or frequencies over which electromagnetic radiation extends. It is the entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light. In the electromagnetic spectrum, the entire distribution of electromagnetic radiation is done according to their frequency or wavelength.
The energy of an electromagnetic wave depends on its frequency and wavelength. The shorter the wavelength, the greater the energy of the electromagnetic wave but the larger frequency, the greater the energy of the electromagnetic wave.
X-rays has a frequency of about 1×10^20 Hz compared to visible light of frequency of about 1×10^15 Hz. Hence X-rays, having a larger frequency, is more energetic than visible light.
Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.
Answer:
a) v = -30 - 32 t , s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s
c) v (1) = -62 ft / s, v (3) = -126 ft / s , d) t = 7.13 s , e) v = -258.16 ft / s
Explanation:
a) For this exercise they give us the function of the position of the ball
s (t) = s (o) + v_o t - 16 t²
notice that you forgot to write the super index
indicate the initial position of the ball
s (o) = 600 ft
also indicates initial speed
v_o = - 30 ft / s
let's substitute in the equation
s (t) = 600 - 30 t -16 t²
to find the speed we use
v = ds / dt
v = v_o - 32 t
v = -30 - 32 t
b) To find the average speed, look for the speed at the beginning and end of the time interval
t = 1 s
v (1) = -30 -32 1
v (1) = - 62 ft / s
t = 3 s
v (3) = -30 -32 3
v (3) = -126 ft / s
the average speed is
v = (v (3) -v (1)) / (3-1)
v = (-126 +62) / 2
v = -32 ft / s
c) instantaneous speeds, we already calculated them
v (1) = -62 ft / s
v (3) = -126 ft / s
d) the time to reach the ground
in this case s = 0
0 = 600 - 30 t -16 t²
t² + 1,875 t - 37.5 = 0
we solve the quadratic equation
t = [-1,875 ±√ (1,875² + 4 37.5)] / 2
t = [1,875 ± 12.39] / 2
t₁ = 7.13 s
t₂ = negative
Since the time must be positive, the correct answer is t = 7.13 s
e) the speed of the ball on reaching the ground
v = -30 - 32 t
v = -30 - 32 7.13
v = -258.16 ft / s
In a Venn diagram, the separate circles contain characteristics unique to each compared and the intersection contains characteristics that are common to both items being compared. This Venn diagram compares the inner and outer planets. What belongs in the center section?
a. -Revolve around the Sun
-Rotate on an axis
-Generally have rings
b. -Revolve around the Sun
-Rotate on axis
-Generally have moons
c. -Rotate around the Sun
-Revolve on an axis
-Generally have moons
d. -Rotate around the Sun
-Revolve on an axis
-Generally have rings
Answer:
B. revolve around the sun
rotate on an axis
generally have moons
Explanation:
edge 2021
g A proton is held at rest in a uniform electric field. When it is released, the proton will lose... electrical potential energy. kinetic energy. both kinetic energy and electric potential energy. neither kinetic energy or electric potential energy.
Answer:
It will lose electrical potential energy.
Explanation:
A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.
Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.
A 1000-kg car is moving down the highway at 14m/s. What is the momentum?
Explanation:
Momentum = mass × velocity
p = mv
p = (1000 kg) (14 m/s)
p = 14000 kg m/s
The momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
Given the data in the question
Mass of the car; [tex]g = 1000kg[/tex]Velocity of the car; [tex]v = 14m/s[/tex]Momentum; [tex]p = ?[/tex]Momentum is the product of the mass of a particle and its velocity.
Momentum = Mass × Velocity
[tex]P = m \ * \ v[/tex]
We substitute our given values into the equation
[tex]P = 1000kg \ * \ 14m/s\\\\P = 14000kg.m/s[/tex]
Therefore, the momentum of the car as it moves down the highway at the given speed is 14000-kg.m/s
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A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
1. An object with a mass of 15 kilograms is pushed by a force of 30 Newtons. How much does
it accelerate?
Answer: [tex]2m/s^2[/tex]
Explanation:
[tex]Formula: F=ma[/tex]
Where;
F = force
m = mass
a = acceleration
Solve for a;
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]
You drive in straight line at 20 m/s for 10 miles, then at 30m/s for an other 10 miles what is your average speed
Answer:
25 m/s
Explanation:
Data provided in the question
20 m/s for 10 minutes
And, the 30 m/s for another 10 minutes
Based on the above information, the average speed is
As we know that
[tex]Average\ speed = \frac{Total\ distance}{Total\ time}[/tex]
[tex]= \frac{20\times10\times60 + 30\times10\times60 }{20\times60}[/tex]
= 25 m/s
1 hour = 60 minutes
1 minute = 60 seconds
Hence, the average speed is 25 m/s
In the question, there are miles is given but instead of this we use the minutes as we have to find out the average speed and time should not be in miles it should be in minutes, hour or seconds
Therefore we considered the same
The air flowing over the top of the wing travels
in the same amount of time than the air
flowing beneath the wing.
Answer: Short Answer: NO ( In Most Cases)
Explanation:
If that were true then planes couldn't get off the ground to fly. The front of the wing is cutting/pushing the air. On the top of the wing the air moves faster and on the bottom it moves slower making a upward draft giving the object the ability to fly or glide.
What is the resistance of a circuit with a voltage of 10 V in a current of 5 A use almond law to create the resistance
Answer:
2Ω
Explanation:
Ohm's law:
V = IR
10 V = (5 A) R
R = 2 Ω
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
What is the momentum of an 8kg bowling ball rolling at 2m/s
Answer:
16kg m/s
Explanation:
P=mv
8 times 2=16kg m/s
Answer:
The momentum of moving body is calculated by
p= mv
In this question m= 8kg
v= 2m/s
so p = 8*2 = 16 kg m/s.
"Mass in motion" can be used to describe momentum. Mass exists in all things. Therefore, if an object is moving, it has momentum—its mass is moving. There are two factors that determine an object's momentum level: how much and how quickly the objects are moving.
Mass and velocity are two variables that affect momentum. An object's momentum can be expressed mathematically as the product of its mass multiply by its velocity.
The equation above can be rewritten as p = m • v, where m is the mass and v is the velocity, since momentum is represented by the lower case p in physics. The equation demonstrates that an object's momentum is directly proportional to its mass and velocity.
The quantity momentum is a vector. A vector quantity is a quantity that is fully described by magnitude and direction, as was discussed in a previous unit. Information about the bowling ball's magnitude as well as its direction must be included in order to fully describe the momentum of a 5-kg ball traveling westward at 2 m/s. The ball has a momentum of 10 kg m/s.
Until information about the ball's direction is provided, the ball's momentum cannot be fully described. The direction of the ball's velocity and the direction of the momentum vector are identical. It was mentioned in a previous unit that the velocity vector moves in the same way that an object moves. The bowling ball's momentum can be fully described as 10 kg m/s westward if it is moving westward. The magnitude and direction of an object's momentum can be used to fully describe it as a vector quantity.
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A scuba diver and her gear displace a volume of 68.5 L and have a total mass of 71.8 kg . Part A What is the buoyant force on the diver in sea water? FB = nothing N Request Answer Part B Will the diver sink or float?
Answer:
A) Fb = 671.3 N
B) The diver will sink.
Explanation:
A)
The buoyant force applied on an object by a fluid is given by the following formula:
Fb = Vρg
where,
Fb = Buoyant Force = ?
V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³
ρ = Density of Water = 1000 kg/m³
g = 9.8 m/s²
Therefore,
Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)
Fb = 671.3 N
B)
Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.
W = mg = (71.8 kg)(9.8 m/s²)
W = 703.64 N
Since, W > Fb. Therefore, the downward force of weight will make the diver sink.
The diver will sink.
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?
Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]
The rms voltage (in V) measured across the secondary coil is calculated as;
[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s 2
Answer:
0.981 rad/sec^2
Explanation:
mass that pulls on string = 5 kg
weight due to mass = 5 x 9.81 = 49.05 N
radius of rod = 0.1 m
torque produced by this force on the rod = force x radius
torque = 49.05 x 0.1 = 4.905 N-m
mass of disk = 125 kg
radius of disk = 0.2 m
moment of inertia of the disk I = m[tex]r^{2}[/tex]
I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2
from the equation, T = Iα
where T is torque
I is moment of inertia
α is angular acceleration
imputing values,
4.905 = 5α
α = 4.905/5 = 0.981 rad/sec^2
Q) Considering the value of ideal gas constant in S.I. unit, find the volume of 35g O2 at 27°C and 72
cm Hg pressure. Later, if we keep this pressure constant, the r.m.s velocity of this oxygen molecules
become double at a certain temperature. Calculate the value of this temperature.
Answer:
V = 0.0283 m³ = 28300 cm³
T₂ = 1200 K
Explanation:
The volume of the gas can be determined by using General Gas Equation:
PV = nRT
where,
P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa
V = Volume of Gas = ?
n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol
R = General Gas Constant = 8.314 J/ mol.k
T = Temperature of Gas = 27°C + 273 = 300 k
Therefore,
(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)
V = 2718.678 J/95992.12 Pa
V = 0.0283 m³ = 28300 cm³
The Kinetic Energy of gas molecule is given as:
K.E = (3/2)(KT)
Also,
K.E = (1/2)(mv²)
Comparing both equations, we get:
(3/2)(KT) = (1/2)(mv²)
v² = 3KT/m
v = √(3KT/m)
where,
v = r.m.s velocity
K = Boltzamn Constant
T = Absolute Temperature
m = mass of gas molecule
At T₁ = 300 K, v = v₁
v₁ = √(3K*300/m)
v₁ = √(900 K/m)
Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?
v₂ = 2v₁ = √(3KT₂/m)
using value of v₁:
2√(900 K/m) = √(3KT₂/m)
4(900) = 3 T₂
T₂ = 1200 K
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________
Answer:
either +z direction or -z direction.
Explanation:
The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.
You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction
In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.
Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,
creating a wave that moves through the crowd and they sit back down again after a while. This wave
move around the stadium without moving the people around it. Considering this situation, justify
your answer about nature of wave.
Answer:
The nature of the wave formed is a transverse progressive wave.
Explanation:
A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.
A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.
The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV
Answer:
1) n = 4.47*10^12 photons
2) K = 0.25 eV
Explanation:
This is a problem about the photoelectric effect.
1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:
[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex] (1)
Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:
[tex]E_p=h\frac{c}{\lambda}[/tex] (2)
c: speed of light = 3*10^8 m/s
h: Planck's constant = 6.626*10^-34 Js
λ: wavelength = 600*10^-9 m
You replace the values of the parameters in the equation (2):
[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]
Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:
[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]
The number of photons is 4.47*10^12
2) The kinetic energy of the electrons emitted by the metal is given by the following formula:
[tex]K=E_p-\Phi[/tex] (3)
Ep: energy of the photons
Φ: work function of the metal = 1.7 eV
You first convert the energy of the photons to eV:
[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]
You replace in the equation (3):
[tex]K=1.95eV-1.7eV=0.25eV[/tex]
The kinetic energy of the electrons emitted by the metal is 0.25 eV
(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]
(2). The maximum kinetic energy of the electron is 0.37eV.
(1). The power of light is given as,
[tex]P=1.4*10^{-6}W[/tex]
Energy is given as,
[tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]
Number of photons per second is,
[tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]
(2). the maximum kinetic energy of the electron is,
[tex]K.E=E-\phi[/tex]
Where [tex]\phi[/tex] is work function.
[tex]K.E=2.07-1.7=0.37eV[/tex]
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A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 40.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.
If the speed of sound is 337 m/s, determine the length of an open tube (open at both ends) that has a fundamental frequency of 233 Hz and a first overtone frequency of 466 Hz.
Answer:
Explanation:
fundamental frequency at closed pipe = 40.4 Hz
overtones are odd harmonics in closed pipe
first three overtones are
3 x 40.4 , 5 x 40.4 , 7 x 40.4 Hz
= 121.2 Hz , 202 Hz , 282.8 Hz .
speed of sound given is 337 , fundamental frequency is 233 Hz
wavelength = velocity of sound / frequency
= 337 / 233
= 1.446 m
for fundamental note in open pipe
wavelength /2 = length of tube
length of tube = 1.446 / 2
= .723 m
= 72.30 cm .
first overtone will be two times the fundamental ie 466. In open pipe all the harmonics are found , ie both odd and even .
A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."
Required:
a. What is the field's magnitude?
b. What is the field's direction?
Answer:
The classification of that same issue in question is characterized below.
Explanation:
The given values are:
Current, I = 50.0 A
Diameter, d = 0.10 cm
(a)...
As we know,
⇒ Magnetic force = Copper wire's weight
So,
⇒ [tex]B\times I\times L=M\times g[/tex]
On putting the estimated values, we get
⇒ [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]
⇒ [tex]50B=69.03297\times 10^{-3}[/tex]
⇒ [tex]B=1.38\times 10^{-3} \ T[/tex]
(b)...
As we know,
⇒ [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]
⇒ [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]
⇒ [tex]=2240\times \pi \ 0.000001[/tex]
⇒ [tex]=7.037\times 10^{-3} \ kg[/tex]
A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.
Answer:
d. 1.00 m
Explanation:
In 1905, Einstein proposed special theory of relativity of light.
This theory had a number of consequences or results. One of them is called "Length Contraction".
According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.
But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.
In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:
d. 1.00 m
The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of Q is located at one of the corners of the square. What is the potential (relative to infinity) at the center when each of the other corners is also contains a point charge of Q
Answer:
12.0 V
Explanation:
Data :
Potential difference due to a single charge (+Q), E = 3.0 V
The Electric potential for the system of charges is given as:
[tex]E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}][/tex]
for single charge, E = 3.0 V = [tex]\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}][/tex] ->eq(1)
And for 4 charges:
[tex]E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}][/tex] -eq(2)
from eq(1) and (2) we have
E = 4 × 3.0 V = 12 V
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
Answer:
[tex]\large \boxed{42\, \mu \text{C}}$[/tex]
Explanation:
The formula for the force exerted between two charges is
[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]
where k is the Coulomb constant.
The charges are identical, so we can write the formula as
[tex]F=k\dfrac{q^{2}}{r^2}[/tex]
[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]
A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.30 kg is attached to one end and a second mass m2 = 3.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
A) What is the moment of inertia of the system?B) If the rod rotates with an angular speed of 2.00 rad/s, how much kinetic energy does the system have?C) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?D) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.00 rad/s?
Answer:
Explanation:
Moment of inertia of the rod = 1/12 m L²
m is mass of the rod and L is its length
= 1/2 x 2.3 x 2 x 2
= 4.6 kg m²
Moment of inertia of masses attached with the rod
= m₁ d² + m₂ d²
m₁ and m₂ are masses attached , and d is their distance from the axis of rotation
= 5.3 x 1² + 3.5 x 1²
= 8.8 kg m²
Total moment of inertia = 13.4 kg m²
B )
Rotational kinetic energy = 1/2 I ω²
I is total moment of inertia and ω is angular velocity
= .5 x 13.4 x 2²
= 26.8 J .
C )
when mass of rod is negligible , moment of inertia will be due to masses only
Total moment of inertia of masses
= 8.8 kg m²
D )
kinetic energy of the system
= .5 x 8.8 x 2²
= 17.6 J .
(A) Total moment of inertia is 13.4 kgm²
(B) Total kinetic energy is 26.8J
(C) Moment of inertia is 8.8 kgm²
(D) Kinetic energy is 17.6J
Rotational motion:
(A) The moment of inertia of the rod is given by:
I = 1/12 mL²
where m is the mass of the rod
and L is the length
So,
I = (1/12) × 2.3 × 2²
I = 4.6 kgm²
Now, the moment of inertia of masses attached to the rod is given by:
I' = m₁ d² + m₂d²
where m₁ and m₂ are masses
and d is their distance from the axis of rotation
I' = 5.3 × 1² + 3.5 × 1²
I' = 8.8 kgm²
The total moment of inertia of the system is given by:
I(tot) = I + I'
I(tot) = 13.4 kgm²
(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:
KE = 1/2 I(tot)ω²
KE = 0.5 × 13.4 × 2²
KE = 26.8J
(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be
I(tot) = I' = 8.8 kgm²
(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:
KE = 1/2 I'ω²
KE = 0.5 × 8.8 × 2²
KE = 17.6J
Learn more about rotational motion:
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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of it as a cord) between the locomotive and the first car (FT1) to that between the first car and the second car (FT2), for any nonzero acceleration of the train
Answer:
The ratio is [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
Here we are assume the acceleration of the train is a
which makes the acceleration of each car a
From the question we are told that
Considering the second car
The force causing it s movement is mathematically represented as
[tex]F_{T2} = ma[/tex]
Considering the first car
The force causing it s movement is mathematically represented as
[tex]F = F_{T1} -F_{T2} = ma[/tex]
=> [tex]F_{T1} -ma = ma[/tex]
=> [tex]F_{T1} = 2 ma[/tex]
=> [tex]\frac{F_{T1}}{ma} = 2[/tex]
=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]
As you get ready for bed, you roll up one of your socks into a tight ball and toss it into the laundry basket across the room. Then, you try to toss the other sock without rolling it up.. What effects whether or not your socks land in the basket?
Answer:
The drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, the size of the sock (weight), and the speed with which the socks is tossed.
Explanation:
The socks, like every other particle or body travelling through air is met by a resistance that impedes its motion. This resistance is due to the air molecules around, that collide with the body as it travels through them. The resistance offered by this force is proportional to the surface area of the body that collides with the air molecule, so, rolling the socks into a ball reduces the effect of air resistance on the socks, compared to the one tossed without rolling. Air resistance is also largely dependent on the relative motion of the body and the air molecules, the density of the fluid (air), and the size of the body (weight).
Therefore, whether the socks lands in the basket or not is affected by the drag (air resistance) it experiences along its flight to the basket, due to the shape and surface area of the socks, size of the socks (weight), and the speed with which the socks is tossed.
Drag force opposes motion of objects through fluid with its magnitude depending on the velocity of the object in the fluid
The single parameter that effects whether or not the socks lands in the basket is the drag force, [tex]\mathbf{F_D}[/tex] acting on the socks
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
The reason that drag force is the parameter that effects the landing point of the socks is as follows:
The parameters that effects whether or not the socks land in the basket or not are;
The distance of the basket away from the thrower = The range, RThe velocity with which the socks are thrown, uThe angle of elevation with which each socks is thrown, θThe amount of drag experienced by each socks, [tex]\mathbf{F_D}[/tex]The parameters, R, u, and θ depends on the thrower, that parameter that effects the whether or not the socks lands in the basket that is independent of the thrower, is the drag, [tex]\mathbf{F_D}[/tex]
Drag is the force opposing (slows) the motion of an object in a fluid.
The drag force, [tex]\mathbf{F_D}[/tex], slowing down motion, is given by the following formula;
[tex]F_D = \mathbf{C_D \times A \times \dfrac{\rho \times v_r^2}{2}}[/tex]
Where;
[tex]v_r[/tex] = The velocity of flow of the fluid, relative to the object
ρ = The density of the fluid
[tex]C_D[/tex] = The drag coefficient
A = The cross sectional area of the fluid
Therefore, the independent parameter that effects whether or not the socks lands in the basket is the drag force on the socks
Learn more about drag force here:
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