Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
an elevator of mass 250kg is carrying two persons whose masses are 50kg and 100kg. if the forces exerted by the motor is 3000N. calculate the mass of the bodies in the elevator.... Taking g as 10m/s²
Explanation:
mass=force*acceleration
mass=3000*10
mass=30,000
The mass of the bodies in the elevator is 400 kg.
The acceleration of the elevator is 2.5 m/s².
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given parameters:
Mass of the elevator: M = 250 kg.
Mass of two persons: m₁ = 50 kg and m₂ = 100 kg.
Force exerted by the motor: F = 3000N.
g = 10 m/s².
Let, the acceleration of the elevator = a.
the mass of the bodies in the elevator :m= 250 kg. + 50 kg +100 kg. = 400 kg.
Now, F = mg - ma
⇒ 3000 = 400×10 - 400a
⇒ a = 1000/400 = 2.5 m/s²
Hence, the acceleration of the elevator is 2.5 m/s².
Learn more about acceleration here:
brainly.com/question/12550364
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To increase the potential energy of the system, what did you have to do?
Answer:
You can use work to add kinetic energy to a system or to increase potential energy in the system.
Explanation:
Potential energy stored in any system can be released as kinetic energy. Kinetic energy can be transformed to do work or to increase potential energy.
hope this helped
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:__________.
Complete question is;
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:
garage door openers: 40.0 MHz
standard cordless phones: 40.0 to 50.0 MHz
baby monitors: 49.0 MHz
FM radio stations: 88.0 to 108 MHz
cell phones: 800 to 900 MHz
Global Positioning System: 1227 to 1575 MHz
microwave ovens: 2450 MHz
wireless internet technology: 2.4 to 2.6 GHz
Which of the following statements correctly describe the various applications listed above? Check all that apply.
a.) All these technologies use radio waves, including low-frequency microwaves.
b.) All these technologies use radio waves, including high-frequency microwaves.
c.) All these technologies use a combination of infrared waves and high-frequency microwaves.
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.
Answer:
B, D, E, F are the correct statements.
Explanation:
Looking at the options;
A) This is true because radio waves are electromagnetic radiation being used today in television, mobile phones, radios and other areas of communication technologies. And the examples given to us fall in the category of technologies that use radio waves.
B) microwaves usually have long wavelengths and low frequencies. However, sometimes they could have high frequencies usually more than radio waves. Thus, this option is correct.
C) This option is wrong because it's not all the listed technologies that use combination of infrared waves and high-frequency microwaves.
D) we are given the frequency of microwave ovens as 2450 MHz.
Converting to GHz gives; 2.45 GHz.
We are told that wireless internet technology has frequency between 2.4 to 2.6 GHz. Thus, microwave frequency falls in the same range as wireless internet technology and thus the statement is true.
E) we know that wavelength is inversely proportional to frequency. This means that the higher the frequency, the shorter the wavelength.
In the frequencies given to us, wireless internet technology have the highest frequency which means they have the shortest wavelength. The statement is true.
F) from the frequencies given to us, the smallest is garage door openers = 40.0 MHz = 40 × 10^(6) Hz while the biggest is 2.6 GHz = 2.6 × 10^(9) Hz
Formula for wavelength is;
Wavelength = speed of light/frequency
Speed of light = 3 × 10^(8) m/s
Thus;
Wavelength = (3 × 10^(8))/(40 × 10^(6))
Or wavelength = (3 × 10^(8))/(2.6 × 10^(9))
So,wavelength = 7.5 m or 0.12 m
This falls into the given range of 0.10 to 10.0 m.
Thus, the statement is true.
What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?
Answer:
[tex]f=4.70\times 10^{14}\ Hz[/tex]
Explanation:
Given that,
The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]
We need to find the frequency of the light. We know that,
[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]
So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].
Which of the following best describes our
atmosphere?
A. envelope of gases that surround Earth
B. a specific range of altitude where plant life flourishes
C. The air, water, and land that form our planet
D. the water vapor in the air surrounding our planet
Easy question just don’t understand it please help.
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
Light rays enter a transparent material. Which description best describes what happens to the light rays
When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.
An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
Which one the answer to this question
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
Explanation:
...
Potential energy is best defined as which of the following?
A Mass energy
B Energy of Motion
C Stored Energy
D Energy of height
Answer: C. Stored energy
The energy goes from _____ to the _____ above it.
Answer:
The energy goes from the ground state to the excited states above it.
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer:
a) The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular speed is 8 seconds.
Explanation:
a) Let suppose that the disk accelerates uniformly, given that initial and final angular speed ([tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex]), in radians per second, and change in angular position ([tex]\Delta \theta[/tex]), in radians, are known. The angular acceleration ([tex]\alpha[/tex]), in radians per square second, are found by using this expression:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex] (1)
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\Delta \theta = 60\,rad[/tex], then the angular acceleration of the disk is:
[tex]\alpha = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \Delta \theta}[/tex]
[tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex]
The angular acceleration is 1.875 radians per square second.
b) The time taken by the disk to reach the final angular velocity is determined by the following kinematic formula:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex] (2)
Where [tex]t[/tex] is the time, in seconds.
If we know that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega_{f} = 15\,\frac{rad}{s}[/tex] and [tex]\alpha = 1.875\,\frac{rad}{s^{2}}[/tex], then the time taken by the disk is:
[tex]t = \frac{\omega_{f}-\omega_{o}}{\alpha}[/tex]
[tex]t = 8\,s[/tex]
The time taken by the disk to reach the final angular speed is 8 seconds.
In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small sphere of clay is dropped onto the disk, and the sphere sticks to the disk. All frictional forces are negligible. What would happened to the angular momentum and the total kinetic energy of the disk-sphere system immediately before and after the collision?
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀)²
Kf = ½ [tex]\frac{I_o^2}{ I_o + m r^2}[/tex] w₀²
we look for the relationship between the kinetic energy
[tex]\frac{K_f}{K_o}[/tex]= [tex]\frac{I_o}{I_o + m r^2}[/tex]
[tex]\frac{K_f}{K_o } < 1[/tex]
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision
For the Earth, most of the information we've learned about its interior, including the mantle and core, comes from drilling down directly into it.
True
False
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answer:
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Explanation:
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A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
Which of the following is NOT something
the atmosphere does?
A. traps in warmth
B. keeps Earth in a spherical shape
C. provides oxygen to breathe
D. protects Earth from meteoroids
Answer:
I think the answer is B, keeps Earth in a spherical shape
The potential energy of a mass-spring system when the spring is fully compressed and the mass is at rest is 200 J. After releasing the mass, assuming there is no dissipative force, the system will oscillate. At a point during the oscillation the potential energy of the system is 50 J. What is the kinetic energy of the mass at that point
Answer:
150 J
Explanation:
If there is not any dissipative force in the system, the mass and the spring will oscillate eternally., but of course, we assume this is a theoretical situation. The conservation of energy in a system implies that the sum of the potential energy plus kinetic energy remains constant, therefore if in the initial point the mass has 200 J (potential energy) and is at rest ( kinetic energy = 0) the overall energy at the beginning is 200 J. At any point of the oscillation if the potential energy is 50 J the kinetic energy must be 150 J.
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
The weight of a column of air pushing
down over an area is called which of
Help Resources
these?
A. density
B. volume
C. mass
D. air
pressure

Answer:
air pressure
:::::::::::::::::
What is the term used to describe the
force pushing on an area or surface?
A. density
B. pressure
C. surface tension
D. volume
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
An airplane flies 1000 miles in 2 hours. What is its average speed in miles per hour?
Answer:
500km per hour
Explanation:
if in 2 hours the airplane flies 1000 km then 1000 divided by 2 is 500km per hour.
What are the five classes of objects that orbit the sun?
A go-cart is traveling at 15 mi/hr. How long does it take the go-cart to travel 3 miles?
Answer:
12 min
Explanation:every 4 minutes is 1 mile