Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answer:
Will be equal to alpha x r; less than UsN
y=k/x, x is halved.
what happens to the value of y
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
what is defect of vision
Answer:
The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?1. Increasing the wavelength of the laser.2. Increasing the distance to the screen.3. Increasing the frequency of the laser.4. Increasing the number of lines per length.
Answer:
Increase in frequency of the laser
Explanation:
Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin
1.) When the acceleration is zero, what can you say about the velocity of an object?
Answer:
it is either constant or zero
Explanation:
If the x-position of a particle is measured with an uncertainty of 1.00×10-10 m, then what is the uncertainty of the momentum in this same direction? (Useful constant: h-bar = 1.05×10-34 Js.)
Answer:
The uncertainty in momentum is 5.25x 10^25Jsm
Explanation:
We know that
h bar = h/2π
So
1.05x 10^34=h/2pπ
h=1.05x 10^ 34(2π)=6.597x 10^-34Js
dp=(6.597x10^-34/4pπ)/(1x10^-10)
=5.25x10^-25 Jsm
Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Answer:
450 x10^-6 T
Explanation:
We know that the magnetic of each wire is derived from
ByB= uoi/2pir
Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)
= 0.2 x 10^ -3T
B2=
4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)
= 0.4 x 10^ -3T
So
(Bnet)² = (Bx)² + ( By)²
= (0.2² + 0.4²)mT
= 450 x10^-6T
The magnitude of magnetic field at the origin is required.
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
x = Location at x axis = 5 cm
y = Location at y axis = 4 cm
[tex]I_x[/tex] = Current at the x axis point = 50 A
[tex]I_y[/tex] = Current at the y axis point = 80 A
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]
Magnitude of the magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
Finding the net magnetic field using the Pythagoras theorem
[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
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The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.
79.70% Mg-24 (23.9872 amu), 10.13% Mg-25 (24.9886 amu), and 10.17% Mg-26 (25.9846 amu).
A. How many neutrons are in Mg-25?
B. What is the average atomic mass of magnesium in these rocks?
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
Answer:
A. number of neutrons of Magnesium Mg = 13
B. The average mass of Mg = 22.29 amu
C. the magnesium composition on Mars is not the same as that on Earth.
Explanation:
Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.
The atomic number of Magnesium Mg = 12
The atomic number of an element is the number of protons present in the atomic nucleus of the element
i.e Atomic number = number of protons = 12
The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.
Mass number = number of protons + number of neutrons
Given that the mass number of Mg = 25
Then;
25 = 12 + number of neutrons
25 - 12 = number of neutrons
13 = number of neutrons
number of neutrons of Magnesium Mg = 13
B. What is the average atomic mass of magnesium in these rocks?
The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.
Therefore the average atomic mass of magnesium can be calculated as:
= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]
= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]
= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]
The average mass of Mg = 22.29 amu
C. Is the magnesium composition on Mars the same as that on Earth? Explain.
The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.
There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks to see the kind of star that our Sun will ultimately become, long, long after it has turned into a white dwarf. Why is the astronomer on duty going to have a bit of a problem satisfying her request? a. All the old stars in our Galaxy are located in globular clusters and all of these are too far away to be seen with the kind of telescope a college or university campus would have. b. After being a white dwarf, the Sun will explode, and there will be nothing left to see. c. The universe is not even old enough to have produced any white dwarfs yet d. Astronomers only let people with PhD's look at these stellar corpses; it's like an initiation rite for those who become astronomers. e. After a white dwarf cools off it becomes too cold and dark to emit visible light
Answer:
b
Explanation:
An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is
[tex]E_{e} = E_{3} - E_{1}[/tex]
E[tex]_{e}[/tex] = initial kinetic energy of the electron
E[tex]_{1}[/tex] = -4 eV
E[tex]_{3}[/tex] = -1 eV
insert the values into the equation above
[tex]E_{e}[/tex] = -1 -(-4) eV
= -1 + 4 = 3 eV
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm
Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]
= [tex]\frac{10+7155}{2650}[/tex]
[tex]\frac{1}{f}[/tex] = [tex]\frac{7165}{2650}[/tex]
⇒ f = [tex]\frac{2650}{7165}[/tex]
= 0.37
The focal length of the eye is 0.37 cm.
A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.
Required:
Calculate the electric energy dissipated in this process.
Answer:
Explanation:
change in magnetic flux = .16 x .16 x .510 - 0
= .013056 weber .
rate of change of flux = change in flux / time
= .013056 / 40.5 x 10⁻³
= .32237
voltage induced = .32237 V
electrical energy dissipated = v² / R where v is voltage , R is resistance
= .32237² / 6.35
= 16.36 x 10⁻³ J .
A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s
(a) The car is undergoing an acceleration of
[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]
so that in 9.02 s, it will have covered a distance of
[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]
The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:
[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]
(b) The wheels have average angular velocity
[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]
where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.
The wheels start at rest, so
[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]
A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?
Answer:
0.03143 Gy
Explanation:
Mass of the human body = 70 kg
Mass of potassium in the human body = 140 g
chemical atomic mass of potassium = 39.1
From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms
Thus,
140g of potassium will contain;
(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms
We are told that the natural abundance of one of the 40K isotopes is 0.012%.
Thus;
Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms
Formula for activity of K-40 is given as;
Activity = (0.693 × number of K-40 atoms)/half life
Activity = (0.693 × 7.2276 × 10^(19))/1300000000
Activity = 3.85 × 10^(10)
We are told that each decay deposits 1.0 MeV of energy into the body.
Thus;
Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV
Now, 1 MeV = 1.602 × 10^(-13) joules
Thus;
Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J
1 Gy = 1 J/kg
Thus;
Yearly dose = 2.25/70 = 0.03143 Gy
PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above
Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Answer:
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric field of the wave is given by = (0.082 V/m) . What is the magnetic vector of the wave at the point P at that instant? (c = 3.0 × 108 m/s)
Answer:
[tex]B=2.74\times 10^{-10}\ T[/tex]
Explanation:
It is given that,
A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m
We need to find the magnetic vector of the wave at the point P at that instant.
The relation between electric field and magnetic field is given by :
[tex]c=\dfrac{E}{B}[/tex]
c is speed of light
B is magnetic field
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T[/tex]
So, the magnetic vector at point P at that instant is [tex]2.74\times 10^{-10}\ T[/tex].
The magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]
The formula relating electric field and the magnetic field is given as;
[tex]c=\frac{E}{B}[/tex]
E is the electric field strengthB is the magnetic vector of the wavec is the speed of lightFrom the formula shown:
[tex]B=\frac{E}{c}\\B=\frac{0.082}{3.0\times 10^8}\\B=2.73 \times 10 ^{-10}T[/tex]
Hence the magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]
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If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Answer:
6000 counts per secondExplanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample
The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1
Answer:
e% = 0.99% this value is within the 5% tolerance given by the manufacturer
Explanation:
Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by
e% = | R_nominal - R_measured | / R_nominal 100
where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors
let's apply this formula to our case
R_nominal = 10 kΩ = 10000 Ω
R_measured = 100 99 Ω
e% = | 10000 - 10099.1 | / 10000 100
e% = 0.99%
this value is within the 5% tolerance given by the manufacturer
A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?
Answer:
counterclockwise
Explanation:
given data
area = 25 cm²
solution
We know that a changing magnetic field induces the current and induced emf is express as
[tex]\epsilon = -N \frac{d \phi }{dt}[/tex] ..................................1
and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field
so when magnetic field decrease and point coming out of the paper.
so induced current in the loop will be counterclockwise
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 2.0 kg determines its density to be 7300 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
The velocity is [tex]v = 2.6*10^{8} \ m/s[/tex]
Explanation:
From the question we are told that
The side of the cube is [tex]l = 0.13 \ m[/tex]
The mass of the object is [tex]m = 2.0 \ kg[/tex]
The density of the object is [tex]\rho = 7300 \ kg / m^3[/tex]
Generally the volume of the object according to the moving observer is mathematically represented as
[tex]V =\frac{m}{\rho}[/tex]
[tex]V =\frac{2}{7300}[/tex]
[tex]V = 2.74*10^{-4} \ m^3[/tex]
Therefore the length of the side as observed by the observer on high speed is mathematically represented as
[tex]L = \sqrt[3]{V}[/tex]
[tex]L = \sqrt[3]{2.74 *10^{-4}}[/tex]
[tex]L =0.065[/tex]
Now the original length of side is mathematically represented as
[tex]L= l * \sqrt{ (1 - ( \frac{ v}{c})^2 )}[/tex]
Where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]v = \sqrt{1 - [\frac{L}{l}]^2} * c[/tex]
=> [tex]v = \sqrt{1 - [\frac{0.065}{0.13}]^2} * c[/tex]
=> [tex]v = 2.6*10^{8} \ m/s[/tex]
Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <
Answer:
1. Energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared >> visible > ultraviolet
Explanation:
Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.
Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:
1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.
i.e energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.
i.e frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.
i.e wavelength: infrared >> visible > ultraviolet
In terms of lowest to the highest energy,frequency, and wavelength is;
Energy: infrared > visible light > ultraviolet
Frequency: infrared > visible light > ultraviolet
Wavelength: ultraviolet > visible light > infrared
The electromagnetic spectrum:
The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.
The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.Learn more about electromagnetic spectrum here:
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A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which a wave travels on this string.
Answer:
Velocity of wave (V) = 5.2 × 10² m/s
Explanation:
Given:
Per unit length mass (U) = 4.80 × 10⁻³ kg/m
Tension (T)= 1,300 N
Find:
Velocity of wave (V)
Computation:
Velocity of wave (V) = √T / U
Velocity of wave (V) = √1300 / 4.80 × 10⁻³
Velocity of wave (V) = √ 270.84 × 10³
Velocity of wave (V) = 5.2 × 10² m/s
A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.
Answer:
24.34 m/s
Explanation:
recall that one of the equations of motions takes the form:
v = u + at
where,
v = final velocity
u = initial velocity (given as 22.2 m/s)
a = acceleration (given as 2.68m/s²)
t = time elapsed during acceleration (given as 0.80s)
since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:
v = u + at
v = 22.2 + (2.68) (0.8)
v = 24.34 m/s
In France, the wall sockets provide an AC voltage with Vrms = 230 V. You want to use an appliance designed to operate in the United States (Vrms = 120 V) and decide to build a transformer to convert the power line voltage in France to the value required by your appliance.
(a) Should you use a "step-down" transformer (to make Vrms smaller) or a "step-up" transformer (which makes Vrms larger)?
a "step-up" transformer
a "step-down" transformer
(b) If the input coil of your transformer has 2760 turns, how many turns should the output coil have?
_____ turns
Answer:
a)step-down" transformer
b) 1440 turns
Explanation:
There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.
In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.
If
Number of turns in primary coil Np= 2760
Number of turns in secondary coil Ns= unknown
Voltage in primary coil Vp= 230 V
Voltage in secondary coil Vs= 120 V
Ns/Np= Vs/VP
NsVp= NpVs
Ns= NpVs/VP = 2760 × 120/230
Ns= 1440 turns
Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood
Answer:
metal conducts heat better than wood.
Explanation:
Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.
