Answer:
0.61 m
Explanation:
The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.
using the relationship
c = fλ
where
c is the speed of light in vacuum = 3 x 10^8 m/s
f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz
λ is the wavelength = ?
λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m
answer to your question is 0.6m
A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.
Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= - 4 m/s
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)
Answer:
5.465 × 10^10 revolutions
Explanation:
Formula for Magnetic Field = m. v/ q . r
M = mass of electron = mass of positron = 9.11 x 10^-31 kg,
radius of the positron = 5.03 mm
We convert to meters.
1000mm = 1m
5.03mm = xm
Cross multiply
x = 5.03/1000mm
x = 0.00503m
q = Electric charge = -1.6 x 10^-19 C
Magnetic field (B) = 0.85 T
Speed of the positron is unknown
0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503
0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v
v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg
v = 6.8408 ×10-22/ 9.11 x 10^-31 kg
v = 750911086.72m/s
Formula for complete revolutions =
Speed × time / Circumference
Time = 2.30s
Circumference of the circular path = 2πr
r =0.00503
Circumference = 2 × π × 0.00503
= 0.0316044221
Revolution = 750911086.72 × 2.30/0.0316044221
= 1727095499.5/0.0316044221
= 546541562294 revolutions
Approximately = 5.465 × 10^10 revolutions
When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface
Answer:
λmax = 372 nm
Explanation:
First we find the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m
c = speed of light = 3 x 10⁸ m/s
h = Planks Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)
E = 8.5 x 10⁻¹⁹ J
Now, from Einstein's Photoelectric Equation:
E = Work Function + Kinetic Energy
8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)
Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J
Work Function = 5.332 x 10⁻¹⁹ J
Since, work function is the minimum amount of energy required to emit electron. Therefore:
Work Function = hc/λmax
λmax = hc/Work Function
where,
λmax = maximum wavelength of light that will produce photoelectrons = ?
Therefore,
λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)
λmax = 3.72 x 10⁻⁷ m
λmax = 372 nm
Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.
Answer and Explanation:
The computation of the shortest wavelength in the series is shown below:-
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
Where
[tex]\lambda[/tex] represents wavelength
R represents Rydberg's constant
[tex]n_f[/tex] represents Final energy states
and [tex]n_i[/tex] represents initial energy states
Now Substitute is
[tex]1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
now we will put the values into the above formula
[tex]= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )[/tex]
[tex]= 1218888.889 m^{-1}[/tex]
Now we will rewrite the answer in the term of [tex]\lambda[/tex]
[tex]\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m[/tex]
So, the whole Paschen series is in the part of the spectrum.
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
[tex]F_{pull}=m_{T}.a[/tex]
[tex]m_{T}=\frac{F_{pull}}{a}[/tex]
[tex]m_{T}=\frac{3615}{1.516}[/tex]
[tex]m_{T}=2384.5[/tex]
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
For [tex]m_{1}[/tex]:
The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
[tex]m_{1} = \frac{F_{12}}{a}[/tex]
[tex]m_{1} = \frac{1387}{1.516}[/tex]
[tex]m_{1}[/tex] = 915kg
For [tex]m_{2}[/tex]:
There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]
[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]
[tex]m_{2}[/tex] = 605kg
For [tex]m_{3}[/tex]:
[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]
[tex]m_{3} = 2384.5-1520.0[/tex]
[tex]m_{3}[/tex] = 864.5kg
A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disk. (we can ignore the small hole at the center). when it is in use, it turns at a consant 220 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?
Answer:
Torque = 0.012 N.m
Explanation:
We are given;
Mass of wheel;m = 750 g = 0.75 kg
Radius of wheel;r = 25 cm = 0.25 m
Final angular velocity; ω_f = 0
Initial angular velocity; ω_i = 220 rpm
Time taken;t = 45 seconds
Converting 220 rpm to rad/s we have;
220 × 2π/60 = 22π/3 rad/s
Equation of rotational motion is;
ω_f = ω_i + αt
Where α is angular acceleration
Making α the subject, we have;
α = (ω_f - ω_i)/t
α = (0 - 22π/3)/45
α = -0.512 rad/s²
The formula for the Moment of inertia is given as;
I = ½mr²
I = (1/2) × 0.75 × 0.25²
I = 0.0234375 kg.m²
Formula for torque is;
Torque = Iα
For α, we will take the absolute value as the negative sign denotes decrease in acceleration.
Thus;
Torque = 0.0234375 × 0.512
Torque = 0.012 N.m
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
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Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance
Answer:
The near point is [tex]n =44.8 \ cm[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 1.50[/tex]
The distance from the eye is [tex]k = 1.8 \ cm[/tex]
The distance of the book from the eye is [tex]z = -28 \ cm[/tex]
Generally the focal length of the glasses is
[tex]f = \frac{1}{P}[/tex]
=> [tex]f = \frac{1}{1.50 }[/tex]
=> [tex]f = 0.667 \ m[/tex]
=> [tex]f = 66.7 \ cm[/tex]
The object distance is evaluated as
[tex]u = z + k[/tex]
=> [tex]u = -28 + 1.8[/tex]
=> [tex]u = -26.2 \ cm[/tex]
The image distance is evaluated from lens formula as
[tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]
=> [tex]v=- \frac{1}{0.0232}[/tex]
=> [tex]v=- 43 \ cm[/tex]
The near point is evaluated as
[tex]n = -v + k[/tex]
=> [tex]n =-(-43) + 1.8[/tex]
=> [tex]n =44.8 \ cm[/tex]
Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m], RD= 6 [m], RE=12 [m].
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
The highest mountain on mars is olympus mons, rising 22000 meters above the martian surface. If we were to throw an object horizontaly off the mountain top, how long would it take to reach the surface? (Ignore atmospheric drag forces and use gMars=3.72m/s^2
a. 2.4 minutes
b. 0.79 minutes
c. 1.8 minutes
d. 3.0 minutes
Answer:
t = 1.81 min , the correct answer is c
Explanation:
This is a missile throwing exercise
The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
the final height is y = 0 and the initial height is y₀ = 22000 m
0 = y₀ + 0 - ½ g t²
t = √y 2y₀ / g
let's calculate
t = √(2 22000 / 3.72)
t = 108.76 s
let's reduce to minutes
t = 108.76 s (1 min / 60 s)
t = 1.81 min
The correct answer is c
When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass
Answer:
This means that mercury has a higher or faster expansion rate than glass
Explanation:
This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).
An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object
Answer:
T_object = 380.35 K
Explanation:
From Stefan–Boltzmann law, the power output is given by the formula:
P = σAT⁴
where;
σ is Stefan-Boltzmann constant
A is area of the radiating surface.
T is temperature of the body
Now, we are told that the power the object emitted is 3 times the power absorbed from the room.
Thus, we have;
P_e = 3P_a
Where P_e is power emitted and P_a is power absorbed.
So, we have;
σA(T_object)⁴ = 3σA (T_room)⁴
σA will cancel out to give;
(T_object)⁴ = 3(T_room)⁴
We are given T_room = 289 K
Thus;
(T_object)⁴ = 3 × 289⁴
(T_object) = ∜(3 × 289⁴)
T_object = 380.35 K
Rank these electromagnetic waves on the basis of their speed (in vacuum). Rank from fastest to slowest.
a. Yellow light
b. FM radio wave
c. Green light
d. X-ray
e. AM radio wave
f. Infrared wave
Answer:
From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):
d. X-ray
c. Green light
a. Yellow light
f. Infrared wave
b. FM radio wave
e. AM radio wave
Explanation:
Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below
V = Fλ
where:\
V = speed (velocity)
F = frequency
λ = wavelength.
From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with the highest to lowest frequencies/ from left to right are:
X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).
This corresponds to the speed from highest to lowest from left to right.
A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?
Answer:
8.1 m
Explanation:
Convert km/h to m/s.
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
Distance = speed × time
d = (12.5 m/s) (0.65 s)
d = 8.125 m
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Answer
Work done is 57.9KJ
Explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ
The following equation is an example of
decay.
181
185
79
Au →
4
2
He+
Answer:
Alp decay.
Explanation:
From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.
A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.
This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.
Therefore, the equation is an example of alpha decay.
Rank the following types of electromagnetic waves by the wavelength of the wave.
a. Microwaves
b. X-rays
c. Radio waves
d. Visible light
Explanation:
In order of Increasing Wavelength of the Electromagnetic Spectrum :
B) X rays
D) Visible light
A) Microwave
C) Radio Waves
Electromagnetic waves in order of decreasing wavelength is X-rays,visible light,microwaves and radio waves.
What are electromagnetic waves?The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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A 1.5 V battery is connected to a 1000 ohm resistor and a 500 ohm resistor in series. The voltage across the 1000 ohm resistor is _____ V.
Answer:
1 volt and 0.5 voltExplanation:
Given data
voltage supplied Vs= 1.5 volts
resistance R1= 1000 ohms
resistance R2= 500 ohms
The total resistance is
Rt= 1000+ 500
Rt= 1500 ohms
The current I is given as
[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]
Voltage across R1
[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]
Voltage across R2
[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]
In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.
A body is thrown vertically upwards with a speed of 95m / s and after 7s it reaches its maximum height. How fast does it reach its maximum height? What was the maximum height reached?
Explanation:
u = 95 m/sec ( Initial speed)
t = 7 sec ( Time of ascent)
According to Equations of Motion :
[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]
Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m
Answer:
332.5 m
Explanation:
At the maximum height, the velocity is 0.
Given:
v₀ = 95 m/s
v = 0 m/s
t = 7 s
Find: Δy
Δy = ½ (v + v₀) t
Δy = ½ (0 m/s + 95 m/s) (7 s)
Δy = 332.5 m
PLEASE HELP FAST WILL GIVE BRAINLIEST The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. 1.experiment 2.hypothesis 3.observation 4.control
The answer is 3. Observation
Explanation:
The sentence "The popcorn kernels popped twice as fast as the last batch" is the result of observing or measuring the time popcorn kernels require to pop. In this context, the sentence best matches the word "observation" which the term used in the Scientific method to refer to statements that are the result of studying a phenomenon, either through the senses such as sight or through precise instruments that allow scientists to understand numerically variables such as time, speed, temperature, etc.
Do an Internet search to determine what minerals are extracted from the ground in order to manufacture the following products:
a. Stainless steel utensils
b. Cat litter
c. Tums brand antacid tablets
d. Lithium batteries
e. Aluminum beverage cans
Answer:
Raw materials are most times gotten from the earth through various forms of extraction procedures.
A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.
B) Cat litter comprises of ceramic products which is made up of clay.
C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.
D)Lithium batteries are made up of elements in the earth such as lithium and carbon.
E)Aluminum beverage cans are made up of aluminum extracted from the ground.
Nuclear plants use radioactive fuel to produce steam which turns a turbine to generate electricity. This is an example of a(n) _____. A) heat pump B) heat mover C) internal combustion engine D) external combustion engine
Answer:
C) internal combustion engineExplanation:
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.65 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
2.2 ft
Explanation:
0.65 lb / 1 ft = 1.4 lb / x
x ≈ 2.2 ft
QUESTION 27
The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of the
plane is 400 degrees C and the linear coefficient of expansion for titanium is 5x10-6/C when flying at 3 times the speed of sound, how
much would a 10-meter long (originally at oC) portion of the airplane expand? Write your final answer in centimeters and show all of your
work.
Answer:
2 cm.
Explanation:
Data obtained from the question include the following:
Original Length (L₁ ) = 10 m
Initial temperature (T₁) = 0°C
Final temperature (T₂) = 400°C
Linear expansivity (α) = 5×10¯⁶ /°C
Increase in length (ΔL) =..?
Next, we shall determine the temperature rise (ΔT).
This can be obtained as follow:
Initial temperature (T₁) = 0°C
Final temperature (T₂) = 400°C
Temperature rise (ΔT) =..?
Temperature rise (ΔT) = T₂ – T₁
Temperature rise (ΔT) = 400 – 0
Temperature rise (ΔT) = 400°C
Thus, we can obtain the increase in length of the airplane by using the following formula as illustrated below:
Linear expansivity (α) = increase in length (ΔL) /Original Length (L₁ ) × Temperature rise (ΔT)
α = ΔL/(L₁ × ΔT)
Original Length (L₁ ) = 10 m
Linear expansivity (α) = 5×10¯⁶ /°C
Temperature rise (ΔT) = 400°C
Increase in length (ΔL) =..?
α = ΔL/(L₁ × ΔT)
5×10¯⁶ = ΔL/(10 × 400)
5×10¯⁶ = ΔL/4000
Cross multiply
ΔL = 5×10¯⁶ × 4000
ΔL = 0.02 m
Converting 0.02 m to cm, we have:
1 m = 100 cm
Therefore, 0.02 m = 0.02 × 100 = 2 cm.
Therefore, the length of the plane will increase by 2 cm.
An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×10-34 Js.)
Answer:
9.82 × [tex]10^{-35}[/tex] Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ = [tex]\frac{h}{mv}[/tex]
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ = [tex]\frac{h}{mv}[/tex]
= [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]
= [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]
= 9.8222 × [tex]10^{-35}[/tex]
The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.
Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon return, the people on earth will have advanced exactly 120 years into the future. According to special relativity, how fast must you travel
Answer:
I must travel with a speed of 2.97 x 10^8 m/s
Explanation:
Sine the spacecraft flies at the same speed in the to and fro distance of the journey, then the time taken will be 6 months plus 6 months
Time that elapses on the spacecraft = 1 year
On earth the people have advanced 120 yrs
According to relativity, the time contraction on the spacecraft is gotten from
[tex]t[/tex] = [tex]t_{0} /\sqrt{1 - \beta ^{2} }[/tex]
where
[tex]t[/tex] is the time that elapses on the spacecraft = 120 years
[tex]t_{0}[/tex] = time here on Earth = 1 year
[tex]\beta[/tex] is the ratio v/c
where
v is the speed of the spacecraft = ?
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
120 = 1/[tex]\sqrt{1 - \beta ^{2} }[/tex]
squaring both sides of the equation, we have
14400 = 1/[tex](1 - \beta ^{2} )[/tex]
14400 - 14400[tex]\beta ^{2}[/tex] = 1
14400 - 1 = 14400[tex]\beta ^{2}[/tex]
14399 = 14400[tex]\beta ^{2}[/tex]
[tex]\beta ^{2}[/tex] = 14399/14400 = 0.99
[tex]\beta = \sqrt{0.99}[/tex] = 0.99
substitute β = v/c
v/c = 0.99
but c = 3 x 10^8 m/s
v = 0.99c = 0.99 x 3 x 10^8 = 2.97 x 10^8 m/s
An average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37∘C37 ∘ C, and the skin is typically 7C∘7C ∘ cooler. By how much does the person’s entropy change per second due to this heat transfer?
Answer:
-4.7 x 10^-3 J/K-s
Explanation:
The Power generated by metabolizing food = 80 W
The watt W is equivalent to the Joules per sec J/s
therefor power = 80 J/s
20% of this energy is not used for heating, amount available for heating is
==> H = 80% of 80 = 0.8 x 80 = 64 J/s
The inner body temperature = 37 °C = 273 + 37 = 310 K
The entropy of this inner body ΔS = ΔH/T
ΔS = 64/310 = 0.2065 J/K-s
The skin temperature is cooler than the inner body by 7 °C
Temperature of the skin = 37 - 7 = 30 °C = 273 + 30 = 303 K
The entropy of the skin = ΔS = ΔH/T
ΔS = 64/303 = 0.2112 J/K-s
change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)
==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s
The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature in degrees Celsius when 1.50 kcal of heat enters 1.50 kg of the following, originally at 15.0°C.(a) water
(b) concrete
(c) steel
(d) mercury
Answer:
Final temperature Water = 20.99-degree celsius
Final temperature Concrete = 24.98 degree celsius
Final temperature Steel = 50.1 degree celsius
Final temperature Mercury = 29.26 degree celsius
Explanation:
Given the mass of each substance = 1.50 kg
Ti = 15
Q = 1.5 kcal = 6276 joule
We have to use the heat capacity of each object so find the heat capacity from the table.
Heat capacity of water = 4186 J/kg degree celsius.
Heat capacity of concrete = 840 J/kg degree celsius.
Heat capacity of steel = 452 J/kg degree celsius.
Heat capacity of mercury = 139 J/kg degree celsius.
Use the below formula to find the final temperature.
[tex]T_f = T_i + \frac{Q}{mc_w} \\[/tex]
[tex]\text{Temperature in the case of water.} \\= 20 + \frac{6276}{1.5 \times 4186 } \\= 20.99 \ degree \ celsius \\\text{Temperature in the case of concrete.} \\= 20 + \frac{6276}{1.5 \times 840 } \\= 24.98 \ degree \ celsius \\\text{Temperature in the case of steel.} \\= 20 + \frac{6276}{1.5 \times 452 } \\= 29.26 \ degree \ celsius \\\text{Temperature in the case of mercury.} \\= 20 + \frac{6276}{1.5 \times 139 } \\= 50.1 \ degree \ celsius \\[/tex]
