Answer:
b) the heavier one will have twice the kinetic energy of the lighter one.
Explanation:
The kinetic energy of object with mass, m
K.E₁ = ¹/₂mv²
where;
m is mass of the object
v is the velocity of the object
Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate
The kinetic energy of object with mass, 2m
K.E₂ = ¹/₂(2m)v²
K.E₂ = 2(¹/₂mv²)
BUT K.E₁ = ¹/₂mv²
K.E₂ = 2(K.E₁)
Therefore, the heavier one will have twice the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
What is the force that attracts objects with mass toward each other?
Explanation:
gravitional force attracts objects with mass toward each other.
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.
Answer:
[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]
Explanation:
The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:
[tex]\Delta \theta = \omega \cdot \Delta t[/tex]
Where:
[tex]\omega[/tex] - Steady angular speed, measured in radians per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:
[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]
[tex]\Delta \theta = 3116.11\,rad[/tex]
The change in angular displacement, measured in revolutions, is given by the following expression:
[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]
[tex]\Delta \theta = 495.944\,rev[/tex]
A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg, walk clockwise around the platform along its edge at the speed of 1.01 m/s with respect to the platform. Your 20.7 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 17.7 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 93.1 kg and radius 1.93 m. Calculate the total angular momentum of the system.
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?
Answer
1.07E22 Joules
Explanation;
We know that mass expands by a factor
=>>1/√[1-(v/c)²]
But v= 0.509c
So
1/√(1 - 0.509²)
=>>> 1/√(1 - 0.2591)
= >> 1/√(0.7409) = 1.16
But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy
And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.
So Multiplying by 119.84
Kinetic energy will be 1.07x 10^22 joules
Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?
Answer:
The width is [tex]w_c = 0.00422 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 6.33*10^{-7} \ m[/tex]
The width of the slit is [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 1.0 \ m[/tex]
Generally the central maximum is mathematically represented as
[tex]w_c = 2 * y[/tex]
Here y is the width of the first order maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]
[tex]y = 0.00211 \ m[/tex]
So
[tex]w_c = 2 *0.00211[/tex]
[tex]w_c = 0.00422 \ m[/tex]
A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
Explanation:
A) Vair = 1.3 L
B) Volume is not reasonable
Explanation:
A)
Assume
m to be total mass of the man
mp be the mass of the man that pulled out of the water
m1 be the mass above the water with the empty lung
m2 be the mass above the water with full lung
wp be the weight that the buoyant force opposes as a result of the air.
Va be the volume of air inside man's lungs
Fb be the buoyant force due to the air in the lung
given;
m = 78.5 kg
m1 = 3.2% × 78.5 = 2.5 kg
m2 = 4.85% × 78.5 = 3.8kg
But, mp = m2- m1
mp = 3.8 - 2.5
mp = 1.3kg
So using
Archimedes principle, the relation for formula for buoyant force as;
Fb = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
Fb = wp = 1.3× 9.81
Fb = 12.7N
But
Fb = (ρ_water × V_air × g)
So
Vair = Fb/(ρ_water × × g)
Vair = 12.7/(1000 × 9.81)
V_air = 1.3 × 10^(-3) m³
convert to litres
1 m³ = 1000 L
Thus;
V_air = 1.3× 10^(-3) × 1000
V_air = 1.3 L
But since the average lung capacity of an adult human being is about 6-7litres of air.
Thus, the calculated lung volume is not reasonable
Explanation:
A bucket filled with water has a mass of 23 Kg and is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. What torque does the weight of water and bucket produce on the cylinder if the cylinder is ont permitted to rotate? (g= 9.8 m/s2)
Answer:
The torque is 11.27 N m
Explanation:
Recall that torque is the vector product of the force times the distance to the pivoting point. So in our case, the distance to the pivoting point is the radius of the cylinder (0.05 m), and the force is given by the weight of the bucket full of water (W = 9.8 * 23 N = 225.4 N)
Then the torque is: 0.05 * 225.4 N m = 11.27 N m
A 384 Hz tuning fork produces standing waves with a wavelength of 0.90 m inside a resonance tube. The speed of sound at experimental conditions is
Answer:
v = 345.6m/s
Explanation:
v = 384 x 0.9 = 345.6
v = 345.6m/s
Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is 4.82×10-11 T.A) Calculate the wavelength.B) Calculate the wave number.C) Calculate the angular frequency.
D) Calculate the electric-field amplitude.
Answer:
A
[tex]\lambda = 361.45 \ m[/tex]
B
[tex]k = 0.01739 \ rad/m[/tex]
C
[tex]w = 5.22 *10^{6} \ rad/s[/tex]
D
[tex]E = 0.01446 \ N/C[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]
The magnetic field amplitude is [tex]B = 4.82*10^{-11} \ T[/tex]
Generally wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]
=> [tex]\lambda = 361.45 \ m[/tex]
Generally the wave number is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]
=> [tex]k = 0.01739 \ rad/m[/tex]
Generally the angular frequency is mathematically represented as
[tex]w = 2 * \pi * f[/tex]
=> [tex]w = 2 * 3.142 * 830*10^{3}[/tex]
=> [tex]w = 5.22 *10^{6} \ rad/s[/tex]
The the electric-field amplitude is mathematically represented as
[tex]E = B * c[/tex]
=> [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]
=> [tex]E = 0.01446 \ N/C[/tex]
This question involves the concepts of wavelength, frequency, wave number, and electric field.
a) The wavelength is "361.44 m".
b) The wave number is "0.0028 m⁻¹".
c) The angular frequency is "5.22 x 10⁶ rad/s".
d) The electric field amplitude is "0.0145 N/C".
a)
The wavelength can be given by the following formula:
[tex]c=f\lambda[/tex]
where,
c = speed of light = 3 x 10⁸ m/s
f = frequency = 830 KHz = 8.3 x 10⁵ Hz
λ = wavelength = ?
Therefore,
[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]
λ = 361.44 m
b)
The wave number can be given by the following formula:
[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]
wave number = 0.0028 m⁻¹
c)
The angular frequency is given as follows:
[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]
ω = 5.22 x 10⁶ rad/s
d)
The electric field amplitude can be given by the following formula:
[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]
E = 0.0145 N/C
Learn more about wavelength and frequency here:
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A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false
Answer:
a) true
Explanation:
The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false
Answer:
The answer is B. falseExplanation:
Current in the same direction
When current flow through to parallel conductors of a given length, when the current flows in the same direction
1. A force of attraction between the wires occurs and this tends to draw the wires inward
2. A magnetic field in the same direction is produced.
Current in opposite direction
when the current is in opposite direction
1. Force of repulsion between the two wires occurs, draws the wire outward
2. A magnetic field in opposite direction occurs
Specular reflection occurs where the light ray in the glass strikes the reflector. If no light is to enter the water, we require that there be reflection only. Which phenomenon prevents the light from entering the water?
Answer:
The critical angle phenomenon.
Explanation:
Critical angle in optics is the smallest angle of incidence of a wave, that will give total reflection of the wave. This phenomenon occurs at the boundary of two medium, where light will normally move from one medium to another.
To prevent light from entering the water, the angle of incidence of the light incident on the water must exceed the critical angle.
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables.
Answer:
HELLO your question has some missing parts below are the missing parts
note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.
--Given Values--
Inlet Temperature: T1 (K) = 325
Inlet pressure: P1 (kPa) = 560
Inlet Velocity: V1 (m/s) = 97
Throat Area: A (cm^2) = 5.3
Pressure upstream of (before) shock: Px (kPa) = 207.2
Mach number at exit: M = 0.1
Answer: A) match number at inlet = 0.2683
B) stagnation temperature at inlet = 329.68 k
C) stagnation pressure = 588.73 kPa
D) ) Throat temperature = 274.73 k
Explanation:
Determining states at several locations in the system
A) match number at inlet
= V1 / C1 = 97/ 261.427 = 0.2683
C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex] = 361.427 m/s
v1 = inlet velocity = 97
B) stagnation temperature at inlet
= T1 + [tex]\frac{V1 ^2}{2Cp}[/tex] = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]
stagnation temperature = 329.68 k
C) stagnation pressure
= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]
Ma = match number at inlet = 0.2683
p1 = inlet pressure = 560
hence stagnation pressure = 588.73 kPa
D) Throat temperature
= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]
Th = throat temperature
T = stagnation temp at inlet = 329.68 k
k = 1.4
make Th subject of the relation
Th = 329.68 * (2 / 2.4 ) = 274.73 k
The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 1/2 in).
If we treat the spring assembly as a single spring, what is the approximate spring constant?
k= ____________
Answer:
The approximate spring constant is [tex]k = 55533.33 \ N/m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 68 \ kg[/tex]
The dip of the car is [tex]x = 1.2 \ cm = 0.012 \ m[/tex]
Generally according to hooks law
[tex]F = k * x[/tex]
here the force F is the weight of the person which is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]m * g = k * x[/tex]
=> [tex]k = \frac{m * g }{x }[/tex]
=> [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]
=> [tex]k = 55533.33 \ N/m[/tex]
Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium
The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -
Answer:
The correct answer are - silver, copper and aluminium all three used as fuse material.
Explanation:
A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.
Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a point r < R1 is:
Answer:
E = 0 r <R₁
Explanation:
If we use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀
in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.
Consequently by Gauss's law the electric field is ZERO
E = 0 r <R₁
At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m
Answer:
a) The value of g at such location is:
[tex]g=9.8005171\,\frac{m}{s^2}[/tex]
b) the period of the pendulum with the length is 0.970 m is:
[tex]T=1.9767 sec[/tex]
Explanation:
Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:
[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]
a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:
[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]
b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:
hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body
Answer:
Acceleration of the body is:
[tex]a=0.27\,\,m/s^2[/tex]
Explanation:
Use Newton's second Law to solve for the acceleration:
[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]
A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?
Answer:
a) 0.67 rad/sec in the clockwise direction.
b) 98.8% of the kinetic energy is lost.
Explanation:
Let us take clockwise angular speed as +ve
For first cylinder
rotational inertia [tex]I[/tex] = 2.0 kg-m^2
angular speed ω = +5.0 rad/s
For second cylinder
rotational inertia [tex]I[/tex] = 1.0 kg-m^2
angular speed = -8.0 rad/s
The rotational momentum of a rotating body is given as = [tex]I[/tex]ω
where [tex]I[/tex] is the rotational inertia
ω is the angular speed
The rotational momenta of the cylinders are:
for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s
for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s
The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s
When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2
Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]
where [tex]I_{t}[/tex] is their total rotational inertia
[tex]w_{f}[/tex] = their final angular speed together
Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]
According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum
this means that
2 = 3[tex]w_{f}[/tex]
[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s
From the first statement, the direction is clockwise
b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]
where [tex]I[/tex] is the rotational inertia
[tex]w[/tex] is the angular speed
The kinetic energy of the cylinders are:
for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J
for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J
Total initial energy of the system = 25 + 32 = 57 J
The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]
where
where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders
[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders
Final kinetic energy = [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J
kinetic energy lost = 57 - 0.67 = 56.33 J
percentage = 56.33/57 x 100% = 98.8%
A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s
B) The percentage of the original kinetic energy lost to friction is;
percentage energy lost = 98.82%
We are given;
Rotational Inertia for first cylinder; I₁ = 2 kg.m²
Angular speed of first cylinder; ω₁ = 5 rad/s
Angular speed of second cylinder; ω₂ = 8 rad/s
Rotational Inertia for second cylinder; I₂ = 1 kg.m²
From conservation of angular momentum, we know that;
Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])
Thus;
I₁ω₁ + I₂ω₂ = I₃ω₃
Where;
ω₃ is the angular speed when the two cylinders are combined
I₃ = I₁ + I₂
I₃ = 2 + 1
I₃ = 3 kg.m²
Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;
(2 * 5) + (1 * -8) = 3ω₃
10 - 8 = 3ω₃
3ω₃ = 2
ω₃ = 2/3
ω₃ = 0.67 rad/s
B) Let us find initial kinetic energy;
E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²
E_i = ¹/₂((2 * 5²) + (1 * 8²)
E_i = 57 J
Final kinetic energy is;
E_f = ¹/₂I₃ω₃²
E_f = ¹/₂ * 3 * 0.67²
E_f = 0.67335 J
Energy lost = 57 - 0.67335 = 56.32665 J
percentage energy lost = (56.32665/57) * 100%
percentage energy lost = 98.82%
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Warm blooded animals are homeothermic; that is, they maintain an approximately constant body temperature. (Forhumans it's about 37 oC.) When they are in an environment that is below their optimum temperature, they use energy derived from chemical reactions within their bodies to warm them up. One of the ways that animals lose energy to their environment is through radiation. Every object emits electromagnetic radiation that depends on its temperature. For very hot objects like the sun, that radiation is visible light. For cooler objects, like a house or a person, that radiation is in the infrared and is invisible. Nonetheless, it still carries energy. Other ways that energy is lost by a warm animal to a cool environment includes conduction (direct touching of a cooler object) and convection (cooler air moving and carrying thermal energy away). See Heat Transfer for a discussion of all three.
For this problem, we'll just consider how much energy an animal needs to burn (obtain from internal chemical reactions) in order to stay warm just from radiation losses. The rate at which an object loses energy through radiation is given by the Stefan-Boltzmann equation:
Rate of energy loss = AεσT4
where T is the absolute (Kelvin) temperature, A is the area of the object, ε is the emissivity (unitless and =1 for a perfect emitter, less for anything else), and σ is the Stefan-Boltzmann constant:
σ = 5.67 x 10-8 J/(s m2 K4)
Consider a patient trying to sleep naked in a cool room (55 oF = 13 oC). Assume that the person being considered is a perfect emitter and absorber of radiation (ε = 1), has a surface area of about 2.5 m2, and a mass of 80 kg.
a. A person emits thermal radiation at a rate corresponding to a temperature of 37 oC and absorbs radiation at a rate (from the air and walls) corresponding to a temperature of 13 oC. Calculate the individual's net rate of energy loss due to radiation (in Watts = Joules/second).
net rate of energy loss = Watts
b. Assume the patient produces no energy to keep warm. If they have a specific heat about equal to that of water (1 Cal/kg-oC) how much would their temperature fall in one hour? (1 Cal = 1kcal = 103 cal)
ΔT = oC
c. Given that the energy density of fat is about 9 Cal/g, how many grams of fat would the person have to utilize to maintain their body temperature in that environment for one hour?
amount of fat needed = g
Answer:
a) 360.7 J/s
b) 16.23 °C
c) 34.48 g
Explanation:
The mass of the person = 80 kg
The person is a perfect emitter, ε = 1
surface area of the person = 2.5 m^2
a) If he emits radiation at 37 °C, [tex]T_{out}[/tex] = 37 + 273 = 310 K
and receives radiation at 13 °C, [tex]T_{in}[/tex] = 13 + 273 = 286 K
Rate of energy loss E = Aεσ([tex]T^{4} _{out}[/tex] - [tex]T^{4} _{in}[/tex] )
where σ = 5.67 x 10^-8 J/(s m^2 K^4)
substituting values, we have
E = 2.5 x 1 x 5.67 x 10^-8 x ([tex]310^{4}[/tex] - [tex]286^{4}[/tex]) = 360.7 J/s
b) If they have specific heat about equal to that of water = 1 Cal/kg-°C
but 1 Cal = 1 kcal = 10^3 cal
specific heat of person is therefore = 10^3 cal/kg-°C
heat loss = 360.7 J/s = 360.7 x 3600 = 1298520 J/hr
heat lost in 1 hour = 1 x 1298520 = 1298520 J
This heat lost = mcΔT
where ΔT is the temperature fall
m is the mass
c is the specific heat equivalent to that of water
the specific heat is then = 10^3 cal/kg-°C
equating, we have
1298520 = 80 x 10^3 x ΔT
1298520 = 80000ΔT
ΔT = 1298520/80000 = 16.23 °C
c) 1298520 J = 1298520/4184 = 310.35 Cal
density of fat = 9 Cal/g
gram of fat = 310.35/9 = 34.48 g
A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?
Answer:
A. 0.0374C
B. 0.012F
C. 18 ohms
Explanation:
See attached file
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube
Answer:
The length of the tube is 85 cm
Explanation:
Given;
speed of sound, v = 340 m/s
first harmonic of open-closed tube is given by;
N----->A , L= λ/₄
λ₁ = 4L
v = Fλ
F = v / λ
F₁ = v/4L
Second harmonic of open-closed tube is given by;
L = N-----N + N-----A, L = (³/₄)λ
[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]
Third harmonic of open-closed tube is given by;
L = N------N + N-----N + N-----A, L = (⁵/₄)λ
[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]
The difference between second harmonic and first harmonic;
[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]
The difference between third harmonic and second harmonic;
[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]
Thus, the difference between successive harmonic of open-closed tube is
v / 2L.
[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]
Therefore, the length of the tube is 85 cm
A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.
Answer:
D.
Inverted and reduced
If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.
A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
Hence option B is correct.
To learn more about the concave mirror refer to the link;
https://brainly.com/question/25937699
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a small rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Answer:
B. mineral oil – density of 0.910 g/mL.
Explanation:
Hello,
In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.
In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.
Best regards.
Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 × 105 m/s. What is the work function, in eV of the metal?
Answer:
12.45eVExplanation:
Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.
Mathematically, KE = hf - Ф where;
h is the Planck constant
f is the frequency = c/λ
c is the speed of light
λ is the wavelength
Ф is the work function
The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;
Ф = hc/λ - KE
Ф = hc/λ - 1/2mv²
Given parameters
c = 3*10⁸m/s
λ = 97*10⁻⁹m
velocity of the electron v = 3.48*10⁵m/s
h = 6.62607015 × 10⁻³⁴
m is the mass of the electron = 9.10938356 × 10⁻³¹kg
Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²
Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²
Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰
Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹
Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷
Ф = 0.1995*10⁻¹⁷Joules
Since 1eV = 1.60218*10⁻¹⁹J
x = 0.1995*10⁻¹⁷Joules
cross multiply
x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹
x = 0.1245*10²
x = 12.45eV
Hence the work function of the metal in eV is 12.45eV
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)
Answer:
a)1984.5nm
b)523mm
Explanation:
A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference
θ=(m+0.5)λ where m= 1,2.3....
Where given from the question the 4th dark Fringe which will take place at m= 3
θ=7/2y
Where y= 567nm
= 7/2(567)=1984.5nm
But
B)tan θ ≈ y/d
And sinθ = mλ/d
y=mλd when m= 1 which is the first bright we have
Then y=(1× 567.D)/d
But the distance from Central to the 4th dark Fringe is 1.83cm then
y= 7λD/2d= 1.83cm
D/d=(2)×(1.83×10^-2)/(7×567×10^-9)
=92221.5
y= (567×10^-9)× (92221.5)
=0.00523m
Therefore, the distance between the first and center is y1-y0= 523mm
