Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
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Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
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the efficiency of a carnot cycle is 1/6.If on reducing the temperature of the sink 75 degrees celcius ,the efficiency becomes 1/3,determine he initial and final temperatures between which the cycle is working.
Answer:
450°C
Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .
η = 1 - T2 / T1
η = 1/6 initially
when T2 is reduced by 65°C then η becomes 1/3
Solution
η = 1/6
1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)
When η = 1/3 :
η = 1 - ( T2 - 75 ) / T1
1/3 = 1 - (T2 - 75)/T1.........................(2)
T2 - T1 = -75 [ because T2 is reduced by 75°C ]
T2 = T1 - 75...........................(3)
Put this in (2) :
> 1/3 = 1 - ( T1 - 75 - 75 ) / T1
> 1/3 = 1 - (T1 - 150 ) /T1
> (T1 - 150) / T1 = 1 - 1/3
> ( T1 -150 ) / T1 = 2/3
> 3 ( T1 - 150 ) = 2 T1
> 3 T1 - 450 = 2 T1
Collecting the like terms
3 T1- 2 T1 = 450
T1 = 450
The temperature initially was 450°C
An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)
Answer:
The induced emf is [tex]\epsilon =7.68 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]
The initial magnetic field is [tex]B_i = 0.14 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2 }[/tex]
substituting values
[tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]
[tex]r = 1.1*10^{-2} \ m[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]
Where [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as
[tex]d\phi = dB * A * cos\theta[/tex]
=> [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]
Here [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field
Also A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]
[tex]A = 3.8 *10^{-4] \ m^2[/tex]
So
[tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]
[tex]d\phi = -5.32*10^{-5} \ weber[/tex]
So
[tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]
[tex]\epsilon =7.68 \ V[/tex]
A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s
Answer:
C) 200 m/s
Explanation:
The sound travels a total distance of 360 m in 0.03 minutes.
v = (360 m) / (0.03 min × 60 s/min)
v = 200 m/s
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After impact, the equipment experiences an acceleration of a = 2kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment.
Answer:
Maximum acceleration is 800m/s^2
Explanation:
See attached file
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:
Answer:
absorbent p = S / c
reflective p = 2S/c
Explanation:
The moment of radiation on a surface is
p = U / c
where U is the energy and c is the speed of light.
In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.
p = S / c
in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process
p = 2S/c
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Answer:
The current is [tex]I = 2042\ A[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 2.2 \ m[/tex]
The radius is [tex]r_i = 30 \ cm = 0.30 \ m[/tex]
The number of turn is [tex]N = 1200 \ turns[/tex]
The magnetic field is [tex]B = 1.4 \ T[/tex]
The magnetic field produced is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{l }[/tex]
making [tex]I[/tex] the subject
[tex]I = \frac{B * l}{\mu_o * N }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
substituting values
[tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]
[tex]I = 2042\ A[/tex]
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center
Answer:
moment of inertia I ≈ 4.0 x 10⁻³ kg.m²
Explanation:
given
point masses = 50g = 0.050kg
note: m₁=m₂=m₃=m₄=50g = 0.050kg
distance, r, from masses to eachother = 20cm = 0.20m
the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by
= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m
moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center
mathematically,
I = ∑m×d²
remember, a square will have 4 equal points
I = ∑m×d² = 4(m×d²)
I = 4 × 0.050 × (14.12 x 10⁻² m)²
I = 0.20 × 1.96 × 10⁻²
I = 3.92 x 10⁻³ kg.m²
I ≈ 4.0 x 10⁻³ kg.m²
attached is the diagram of the equation
4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
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A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.
Answer:
2.5×10¹⁹
Explanation:
4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
Someone help find centripetal acceleration plus centripetal force!
Answer:Centripetal force that acts an object keep it along a moving circular path.
Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.
a=v/r
object will along moving continue a straight path unless by the external force.External force is the centripetal force.
Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.
Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared
(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.
The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.
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3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
O A. Ferromagnetic
O B. Non-magnetic
C. Diamagnetic
O D. Paramagnetic
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
Answer:
its paramagnetic
Explanation:
i took this quiz
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ............ cm
(b) the time interval required to complete one revolution ............ s
Answer:
(a) 3.9cm
(b) 1.66 x 10⁻⁸s
Explanation:
Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,
F = m x a --------------(i)
Where;
m = mass of the particle
a = acceleration of the mass
The centripetal acceleration is given by;
a = v² / r [v = linear velocity of particle, r = radius of circular path]
Therefore, equation (i) becomes;
F = m v²/ r --------------------(ii)
The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;
F = qvBsinθ -------------(iii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = angle between the velocity and the magnetic field
Combine equations (ii) and (iii) as follows;
m (v² / r) = qvBsinθ [divide both side by v]
m v / r = qBsinθ [make r subject of the formula]
r = (m v) / (qBsinθ) ---------(iv)
(a) From the question;
v = 1.48 x 10⁷m/s
B = 2.14mT = 2.14 x 10⁻³T
θ = 90° [since the direction of velocity is perpendicular to magnetic field]
m = mass of electron = 9.11 x 10⁻³¹kg
q = charge of electron = 1.6 x 10⁻¹⁹C
Substitute these values into equation (iv) as follows;
r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)
r = 3.9 x 10⁻²m
r = 3.9cm
Therefore, the radius of the circular path is 3.9cm
(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by
T = d / v --------------(*)
Where;
d = distance traveled in the circular path in one complete turn = 2πr
v = velocity of the motion = 1.48 x 10⁷m/s
d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]
d = 2(3.142)(3.9 x 10⁻²) = 0.245m
Substitute the values of d and v into equation (*) as follows;
T = 0.245 / 1.48 x 10⁷
T = 0.166 x 10⁻⁷s
T = 1.66 x 10⁻⁸s
Therefore, the time interval is 1.66 x 10⁻⁸s
"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."
Answer:
9.6×10⁹ A
Explanation:
From the question above,
P = VI.................... Equation 1
Where P = Electric power, V = Voltage, I = current.
make I the subject of the equation
I = P/V............. Equation 2
Given: P = 200 kW = 200×10³ W, V = 48000 V.
Substitute these vales into equation 2
I = 200×10³×48000
I = 9.6×10⁹ A.
Hence the current in the line is 9.6×10⁹ A.
A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?
Answer:
Body inertia I = 4.5 kg/m^2
Explanation:
Here, we want to calculate the body inertia when the arms are stretched outwards.
We know from the question that angular momentum is conserved
Thus;
I * 3 = 4.5 * 3
I = 4.5 kg/m^2
A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)
Answer:
Power=50.17dioptre
Power=50.17D
Explanation:
P=1/f = 1/d₀ + 1/d₁
Where d₀ = the eye's lens and the object distance= 5.70m=
d₁= the eye's lens and the image distance= 0.02m
f= focal length of the lense of the eye
We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m
Therefore, we can calculate the power using above formula
P= 1/5.70 + 1/0.02
Power=50.17dioptre
Therefore, the power the eye's is using to see the object from distance is 5.70D
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Answer:
a) 6738.27 J
b) 61.908 J
c) [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] = [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
[tex](I_{1} +I_{2} )w[/tex]
where the subscripts 1 and 2 indicates the values first and second flywheels
[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
where m is the mass of the car
[tex]v_{car}[/tex] is the velocity of the car
Equating the energy
2246.09 = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
making m the subject of the formula
mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.
Work done on the skateboard
a. Rodger Work= 0J
b. Rodger work= 1300J
c. sally work= 1300J
d. sally work= 5850J
e. rodger work= 5850J
Answer:
b. Rodger work = 1300 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of the force.
From the question,
Work is done by Rodger using a force of 100 N in pushing the skateboard through a distance of 13.0 m.
W = F×d............. Equation 1
Where W = work done, F = force, d = distance.
Given: F = 100 N, d = 13 m
Substitute these values into equation 1
W = 100(13)
W = 1300 J.
Hence the right option is b. Rodger work = 1300 J
g a conductor consists of an infinite number of adjacent wires, each infinitely long. If there are n wires per unit length, what is the magnitude of B~
Answer:
B=uonI/2
Explanation:
See attached file
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s