Answer:
E.l soni.do es un.a sensac.ión, en el órg.ano del oído, prod.ucida por el movimie/nto ondu>latorio de un m/edio elástico (normal/mente el aire), debi.do a ra.pidísimos ca/mbios de pre.sión, generado/s por el movimiento vibrat.orio d.e un cuerpo sonoro. ... /El ruido se consid/era a to/do sonid.o / o no de.seado.
Explanation:
2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time
Answer:
(c) at rest
Explanation:
Given
See attachment for the distance time graph
Required
What does the graph illustrate?
From the graph, we can see that the line of distance is a horizontal line.
This suggests that a time increases, the distance remains unchanged
When distance remains unchanged over time, then it means the object is at rest.
Hence, (c) is correct
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth
Answer:
0 N
Explanation:
This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N
What is binding energy?
A.' The attractive forces between the protons in the nucleus and the
electrons
B. The energy required to force two nuclei to undergo nuclear fusion
C. The amount of energy stored in the strong nuclear forces of the
nucleus
D. The amount of energy required to overcome an activation energy
barrier
Please help me out.
Answer:
the answer is B i hope it helps :)
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
B. The energy required to force two nuclei to undergo nuclear fusion. ✅
They are usually expressed in terms of [tex]\sf\purple{kJ/mole}[/tex] of nuclei or [tex]\sf\pink{MeV's/nucleon}[/tex].[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts
Answer:
C. 6.9 watts
Explanation:
Power = work/time
if work = force×distance...
Then... power= (force×distance)/time
Power = (20×12)/35
= 6.9 watts
Kilometer is a unit of length where as kilogram is a unit of mass
By George, you've nailed it, Stacy !
That's a fact, uh huh.
Truer words were never written.
Your statement is one of unquestionable veracity.
The pure truthiness of it cannot be denied.
Was there a question you wanted to ask ?
A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.
Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?
Answer:
a) μ = 0.0136, b) F = 22.8 N
Explanation:
This exercise must be solved in parts. Let's start by using conservation of moment.
a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved
initial instant. Before the crash
p₀ = m v₀
final instant. After inelastic shock
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m + M} \ v_o[/tex]
We look for the speed of the block with the bullet inside
v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]
v = 0.448 m / s
Now we use the relationship between work and kinetic energy for the block with the bullet
in this journey the force that acts is the friction
W = ΔK
W = ½ (m + M) [tex]v_f^2[/tex] - ½ (m + M) v₀²
the final speed of the block is zero
the work between the friction force and the displacement is negative, because the friction always opposes the displacement
W = - fr x
we substitute
- fr x = 0 - ½ (m + M) vo²
fr = ½ (m + M) v₀² / x
the friction force is
fr = μ N
μ = fr / N
equilibrium condition
N - W = 0
N = W
N = (m + M) g
we substitute
μ = ½ v₀² / x g
we calculate
μ = ½ 0.448 ^ 2 / 0.75 9.8
μ = 0.0136
b) Let's use the relationship between work and the variation of the kinetic energy of the block
W = ΔK
initial block velocity is zero vo = 0
F x₁ = ½ M v² - 0
F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]
F = ½ 2.50 0.448² / 0.0110
F = 22.8 N
FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN
From the water whole water cycle starts again.
Most possibly water should be the answer.Which type of energy is stored in a battery?
A. Nuclear energy
B. Electromagnetic energy
C. Chemical energy
D. Electrical energy
SUBMI
Answer:
c
Explanation:
in food and batteries chemical energy is stored :) hope this helped
please help very easy 5th grade work giving brainliest
Answer:
the answer is option B because opposit sides of the magnets attract each other
a 1600 kg car rounds a curve of radius 71 m banked at an angle of 15, What is the magnitude of the friction force required for the car to travel at 86 km/h
Answer:
The frictional force required for the car to travel is 8,365.01 N
Explanation:
Given;
mass of the car, m = 1600 kg
radius of the curved road, r = 71 m
banking angle, θ = 15⁰
velocity of the car, v = 86 km/h = 86/3.6 = 23.89 m/s
The two forces acting on the are:
1. the parallel force to the banked plane
2. the centripetal force pushing the car up the banked plane
To keep the car traveling at 86 km/h;
frictional force + parallel force to the plane = centripetal force pushing the car up the banked plane
The parallel force to the banked plane:
F = mgsinθ
F = 1600 x 9.8 x sin(15⁰)
F = 4,057.98 N
The centripetal force pushing the car up the banked plane:
[tex]F_c= (\frac{mv^2}{r} )cos(\theta)\\\\F_c = (\frac{1600 \times 23.89^2}{71} )cos(15^0)\\\\F_c = 12,422.99 \ N[/tex]
The frictional force required for the car to travel:
[tex]F_k = F_c - F\\\\F_k = 12,422.99 \ N - 4,057.98 \ N\\\\F_k = 8,365.01 \ N[/tex]
Therefore, the frictional force required for the car to travel is 8,365.01 N
Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?
Answer:
Freezing Point - Lower
Boiling Point - Higher
Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
Learn more about average force of friction : https://brainly.com/question/16207943
Your question has some missing data below are the missing data related to your question
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack
of groceries?
the answer is 396 joules :D
From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height of the tower is 200 m, how long will it take for the ball to hit the ground? Ignore air resistance.
Answer:
66,7 seconds
Explanation:
the formula for height/distance is : S=v.t
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt
Answer:
a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]
b) attached below
c) 0.037 m/s
Explanation:
a) Determine the magnetic "drag" force acting at the moment
speed = v
first step: determine current in the loop
I = [tex]\frac{\pi d^2}{16pl} B lv[/tex] ----- ( 1 )
given that the current will induce force on the three sides of the loop found in the magnetic field
forces on vertical sides = + opposite
we will cancel out
hence equation 1 becomes
F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex] ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )
b) Determine the formula for Vt
attached below
c) Find Vt
given :
B = 0.80 T
density of copper = 8.9 * 10^3 kg/m^3
resistivity of copper = 1.68 * 10^-8 Ωm
∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2
= 0.037 m/s
PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 677 nm
B) 218 nm
C) 345 nm
D) 428 nm
OPTION D is the correct answer.
Refer to the attachment for complete calculation...
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?
Answer:
The total momentum of the cars before the collision is 61,000 kg.m/s
The total momentum of the cars after the collision is 61,000 kg.m/s
The velocity of the cars after the collision is 27.727 m/s
Explanation:
Given;
mass of the first car, m₁ = 1000 kg
initial velocity of the car, u₁ = 25 m/s
mass of the second car, m₂ = 1200 kg
initial velocity of the second car, u₂ = 30 m/s
The common velocity of the cars after collision = v
The total momentum of the cars before collision is calculated as;
P₁ = m₁u₁ + m₂u₂
P₁ = (1000 x 25) + (1200 x 30)
P₁ = 61,000 kg.m/s
The total momentum of the cars after collision is calculated as;
P₂ = m₁v + m₂v
where;
v is the common velocities of the cars after collision since they stick together.
P₂ = v(m₁ + m₂)
To determine "v" apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(1000 x 25) + (1200 x 30) = v(1000 + 1200)
61,000 = 2,200v
v = 61,000/2,200
v = 27.727 m/s
The total momentum after collsion = v(m₁ + m₂)
= 27.727(1000 + 1200)
= 61,000 kg.m/s
Thus, momentum before and after collsion are equal.
how do you use the coefficient to calculate the number of atoms in each molecule?
Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS
For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
Did you add the links of the websites you used for this project to the Student Worksheet?
Did you use the scaled sizes to create models of the Sun and the planets?
Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
Did you label each component of your model?
Did you add an attention-catching title above or below your model?
Did you write your name on the back of your poster board?
Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.
Answer
I hope this help....
Explanation:
Answer:
Hope this helps
Explanation:
Calculating Acceleration
Initial
velocity
Time to travel
0.25 m
Final
velocity
Acceleration
Time to travel
0.50 m
# of
washers
11
(m/s)
V2
(m/s)
ti
(s)
t₂
(s)
a = (v2 - v4)/(t2-tı)
(m/s)
1
0.11
0.28
2.23
3.13
0.19
2
0.13
0.36
1.92
2.61
The acceleration of the car with two washers added to the string would be
I can not even read this question.
What are you trying to even say?
The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0.13 m/s.Final velocity = 0.36 m/s.Initial time = 1.92 seconds.Final time = 2.61 seconds.What is an acceleration?An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.
How to calculate average acceleration.In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.
Mathematically, average acceleration is given by this formula:
[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]
Where:
V is the final velocity.U is the initial velocity.[tex]t_i[/tex]initial time measured in seconds.[tex]t_f[/tex] final time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]
a = 0.33 [tex]m/s^2[/tex]
Read more on acceleration here: brainly.com/question/24728358
The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?
[pleas ee helpppp)
I= 0.39 A
OPTION B is the correct answer.
Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms
Answer:
Sorry I don't know the answer
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.
Answer:
564
Explanation:
Answer in your PE notebook
I have learned that
I have realized that
I will apply
Answer:
physical science is important
hety
civil engineering
Lighting is the movement of?
Explanation:
Movement:refers to the changing in the lights whether it be a change in intensity, color or direction of origin.
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
Answer:
For (a): The chemical shift is [tex]2.08\delta[/tex]
For (b): The chemical shift is [tex]9.85\delta[/tex]
For (c): The chemical shift is [tex]7.5\delta[/tex]
Explanation:
To calculate the chemical shift, we use the equation:
[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]
Given value of spectrometer frequency = 200 MHz
For (a):Given peak position = 416 Hz
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]
For (b):Given peak position = [tex]1.97\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]
For (c):Given peak position = [tex]1.50\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
A) 1.67 x 10^-4 s^-1
B) 5.43 x 10^-4 s^-1
C) 1.40 x 10^-4 s^-1
D) 2.22 x 10^-4 s^-1
OPTION C is the correct answer.
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface
Answer:
Shiny surface.
Explanation:
We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.
Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.
This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,
