Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
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
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 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 .
Which symbol is used to show vector quantities
Answer: arrows
Explanation:
A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.
What is the vector quantity unit?
The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.
The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.
Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).
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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.
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.
Answer:
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
Explanation:
The Free Body Diagram associated with the experiment is presented as attachment included below.
Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.
Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:
[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]
Where:
[tex]W[/tex] - Weight of the eraser, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]\theta[/tex] - Angle of the incline, measured in degrees.
The maximum allowable static friction force is:
[tex]f = \mu_{s} \cdot N[/tex]
Where:
[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
Likewise, the kinetic friction force is described by the following model:
[tex]f = \mu_{k} \cdot N[/tex]
Where:
[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:
[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]
[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]
But [tex]N = m\cdot g \cos \theta[/tex], so that:
[tex]\mu = \tan \theta[/tex]
Now, coefficients of static and kinetic friction are, respectively:
[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]
[tex]\mu_{s} \approx 0.716[/tex]
[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]
[tex]\mu_{k} \approx 0.596[/tex]
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
What percent of our solar system's mass is in the sun?
Answer:
99.8
Explanation:
most massive the sun is at the center of the universe
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 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
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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What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT magnetic field if the maximum value of the induced emf is to equal 8.0 V
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.
Calculation of the minimum frequency:Since
number of turns, N = 200 turns
cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
the magnitude of magnetic field strength, B = 30 x 10⁻³ T
the maximum value of the induced emf, E = 8 V
Now
Maximum induced emf should be
E = NBAω
here,
ω is angular velocity (ω = 2πf)
Now
E = NBA2πf
here,
f is the minimum frequency
So,
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz.
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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 tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
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 .
We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
A gas in a closed container is heated with 12J of energy, causing the lid of the container to rise 3m with 5N of force. What is the total change in energy?
Answer:
27J
Explanation:
From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.
Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;
U = Q + (F×d);
U- change in internal energy
Q is the energy received by the system and is positive when energy is received by the system.
Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.
U=12+(3×5)= 27J
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
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A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
[tex]37.8=\frac{1}{2}(3)v^2[/tex]
[tex]v^2=\frac{37.8\times 2}{3}[/tex]
[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]
v=5.02 m/s
c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]
[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/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
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A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)
Required:
a. Qualitatively, indicate the polarization of the plastic.
1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.
b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.
1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.
c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.
1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.
Answer:
(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center
(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.
Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.
(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.40 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
[tex]vf^2 = vi^2 - 2*g*x[/tex]
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
[tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
[tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
A 1KW electric heater is switched on for ten minutes
How much heat does it produce?
Explanation:
P=W/T ==> 1000w = Q/600 ==> Q=600000j
If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .
What is thermal energy ?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .
As given in the problem , we have to find out the heat produced by the 1 - kilo watts electric heater if it is switched on for ten minutes ,
The heat produced by the electric heater = Power × time
= 1000 × 600 Joules
= 600 kilo - Joules
Thus , the heat produced by the electric heater would be 600 - kilo Joules .
To learn more about thermal energy here , refer to the link ;
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An experiment invilves three charges objects: A, B, and C. Object A repels object B and attracts onject C. object C ir repelled by ebonite charged with fur. What is the charge on the object?
Answer:
A and B is positive charge
C_negative
Explanation:
because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract
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]
A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)
Answer:
Its initial position was 471 m.
Explanation:
We have,
Final position of the object is 327 m
Displacement of the object is -144 m
It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,
[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]
So, its initial position was 471 m.
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.Part aIf the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.part bFind Pnet, the net power radiated by the person when in a room with temperature Troom=20∘C
Answer:
The thermal power emitted by the body is [tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated is [tex]P_{net} = 460 \ W[/tex]
Explanation:
From the question we are told that
The length of the assumed hum[tex]T_{room} = 20 ^oC[/tex]an body is L = 2.0 m
The circumference of the assumed human body is [tex]C = 0.8 \ m[/tex]
The Stefan-Boltzmann constant is [tex]\sigma = 5.67 * 10^{-8 } \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
The temperature of skin [tex]T_{body} = 30^oC[/tex]
The temperature of the room is
The emissivity is e=0.6
The thermal power radiated by the body is mathematically represented as
[tex]P_t = e * \sigma * T_{body}^4[/tex]
substituting value
[tex]P_t = 0.6 * 5.67*10^{-8} * (303)^4[/tex]
[tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated by the body is mathematically evaluated as
[tex]P_{net} = P_t * A[/tex]
Where A is the surface area of the body which is mathematically evaluated as
[tex]A = C* L[/tex]
substituting values
[tex]A = 0.8 * 2[/tex]
[tex]A = 1.6 m^2[/tex]
=> [tex]P_{net} = 286.8 * 1.6[/tex]
=> [tex]P_{net} = 460 \ W[/tex]
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
An electric point charge of Q = 22.5 nC is placed at the center of a cube with a side length of a = 16.3 cm. The cube in this question is only a mathematical object, it is not made out of any physical material. What is the electric flux through all six sides of the cube?
Answer:
The electric flux is [tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]
Explanation:
From the question we are told that
The magnitude of the electric point charge [tex]q = 22.5 nC = 22.5 *10^{-9} \ C[/tex]
The length of the one side of the cube is [tex]l = 16.3 \ cm = 0.163 \ m[/tex]
The number of sides is [tex]N= 6[/tex]
The electric flux according to Gauss law is mathematically evaluated as
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Where [tex]\epsilon _ o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12}\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]\phi = \frac{22.5 *10^{-9}}{8.85 *10^{-12}}[/tex]
[tex]\phi = 2.5 *10^{3} \ Nm^2 \cdot C^{-1}[/tex]