Nuclear energy
Electromagnetic energy
Energy transformation/conversion
Energy
Explanation:The energy released in a nuclear reaction( nuclear fusion/fission) is called nuclear energy
Electromagnetic energy is a radiant energy that travels through space in form of electric and magnetic waves at the speed of light.
Therefore, energy, such as sunlight, that travels across space or through matter as electric or magnetic waves is called electromagnetic energy
The process in which energy changes from one type of form to another is called energy transformation or energy conversion
Energy is the ability to do work or cause changes in matter
Dierks is a scientist who is currently focusing on developing a ceramic that holds heat well. He is trying to develop the ceramic to make a higher-temperature curling iron. The application of solid-state physics Dierks is most likely involved in is _______ ……..i just need help understanding the basis behind this question I don’t really understand it
Dierks is most likely involved in is Tt=hermal
How far will a rock starting from rest fall freely in 6.00 seconds
Given data
*The initial speed of the rock is u = 0 m/s
*The given time is t = 6.00 s
*The value of the acceleration due to gravity is g = -9.8 m/s^2
The formula for the distance traveled by the rock is given by the equation of motion as
[tex]d=ut+\frac{1}{2}gt^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} d=(0)(6.0)+\frac{1}{2}(-9.8)(6.0)^2 \\ =-176.4\text{ m} \\ \approx-176\text{ m} \end{gathered}[/tex]What is the speed of a 16 cm wave with a period of 8 seconds?
Given:
The wavelength is,
[tex]\lambda=16\text{ cm}[/tex]The period is,
[tex]T=8\text{ s}[/tex]To find:
The speed of the wave
Explanation:
The speed of the wave is,
[tex]v=\frac{\lambda}{T}[/tex]Substituting the values we get,
[tex]\begin{gathered} v=\frac{16}{8} \\ =2\text{ cm/s} \\ =0.02\text{ m/s} \end{gathered}[/tex]Hence, the speed is 0.02 m/s.
In an internet video an athlete launches a basketball from a stadium platform that is 16.2m higher than the hoop. He makes the basket by launching the ball at an angle of 14.5 degrees above the horizontal with a speed of 11.6 m/s. What horizontal distance does the ball travel before passing through the hoop?
First, lets draw a diagram of the situation
From what we are given, we know couple of things
First:
We know that the net vertical distance travelled is 16.2 meters down
inital velocity is 16.2 meters
And that the acceleration of the ball downwards is 9.8 m/s^2
Lets start by working backwards and figuring out what we need
Using kinematics, we know that the formula we need to find the horizontal distance is
[tex]\Delta x=v_xt[/tex]where x is the distance traveled, vx is the horizontal speed and t is time
We can figure vx right now by finding the horizontal component of speed.
[tex]v_x=v_0cos(14.5)[/tex]which yields us vx = 11.23 m/s
Now we have to find time traveled.
For this, we will need to use the vertical components
Since we have acceleration, inital speed, and distance traveled and we are finding time, we can use the formula
[tex]\Delta y=v_yt+\frac{1}{2}at^2[/tex]To find vy, we just use the same formula we used for vx except we use sin instead of cos. This yields us 2.9 m/s for initial vertical speed.
Plugging in our values, we have
-16.2 = (2.9)t + 1/2(-9.8)(t^2)
To solve for t, we can use the quadratic formula
1/2(-9.8)t^2 + 2.9t + 16.2 = 0
-4.9t^2 + 2.9t + 16.2 = 0
The quadratic formula is as shown
[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Where t is what we solve for, a is the coefficient for the first term (the t^2), b is the coefficient for the second term (t), and c is the coefficient for the last term (16.2).
Plugging these values in (a = -4.9, b = 2.9, c = 16.2) gets us t
[tex]t=\frac{-2.9\pm\sqrt{(2.9)^2-4(-4.9)(16.2)}}{2(-4.9)}[/tex][tex]t=\frac{2.9\pm\sqrt{8.41+317.52}}{9.8}[/tex][tex]t=\frac{2.9\pm18.054}{9.8}[/tex][tex]t=\frac{20.95}{9.8},t=\frac{-15.15}{9.8}[/tex][tex]t=\lbrace2.14\rbrace-1.55[/tex]We solve for t, which gets us 2.14 seconds (we don't use the negative value since we start at t = 0 and time moves forward).
now that we know both time and horizontal speed, we can multiply both of them together to get horizontal distance.
[tex]\Delta x=v_xt[/tex][tex]\Delta x=11.23(2.14)[/tex]This equals 24.012 meters traveled
Light passes from material A, which has an index of refraction of 4/3, into a material B, which has an index of refraction of 5/4. Find the ratio of the speed of light in the material B to the speed of light in material A.
Given,
Light passes from material A, which has an index of refraction of 4/3 into material B, which has an index of refraction of 5/4.
To find: Find the ratio of the speed of light in material B to the speed of light in material A.
Solution:
The refractive index is inversely proportional to the speed of the light in the material.
[tex]\begin{gathered} \frac{n_A}{n_B}=\frac{v_B}{v_A} \\ \frac{\frac{4}{3}}{\frac{5}{4}}=\frac{v_B}{v_A} \\ \frac{v_{B}}{v_{A}}=\frac{16}{15} \end{gathered}[/tex]Thus, the ratio of the speed of light in the material B to the speed of light in material A is
[tex]\frac{v_B}{v_A}=\frac{16}{15}[/tex]Which of the following is a large source of biomass?Question 14 options:Sewage sludgeUraniumClean coalRadioactive waste
We are asked to determine which of the given options is a source of biomass.
Recall that biomass is plant or animal based waste that can be used for energy generation.
Clean coal is a fossil fuel, not biomass.
Uranium is the source of nuclear energy and Radioactive waste is the hazardous waste released as a result. Hence, it is not a biomass source either.
Sewage sludge is the source of biomass.
Therefore, sewage sludge is the correct answer.
Your friend is up next at the bowling alley. Suppose when they take their turn, the ball collided with the same pin, but this time the collision is a glancing collision. After the collision, the pin scatters at an angle of 12 degrees below the horizontal while the ball scatters at an angle of 6 degrees above the horizontal. What is the magnitude of the velocity of the bowling ball and bowling pin if the mass of the pin is 0.68 kg, the mass of the bowling ball is m= 1.6 kg, and the initial velocity is 6m/s.
For this problem, you have to set two equations that maintain constant the momentum in x and y axis.
[tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Px=(6m/s)(1.6kg)=(V1\cdot cos(6))(1.6kg)+(V2\cdot cos(12))(0^.68kg) \\ \sum_{n\mathop{=}0}^{\infty}Py=0=(V1\cdot sin(6))(1.6kg)-(V2\cdot sin(12))(0.68kg) \end{gathered}[/tex][tex]\begin{gathered} 9.6=(V1\cdot1.59)+(V2\cdot0.665) \\ 0=(V1\cdot0.167)-(V2\cdot0.141) \\ \\ V1=V2\cdot0.847 \end{gathered}[/tex]Then you find the relation between the two values and solve the equations system.
[tex]\begin{gathered} 9.6=V2\cdot0.847\cdot1.59+V2\cdot0.665 \\ 9.6=V2(1.346+0.665)=V2\cdot2.011 \\ V2=4.77m/s,\text{ the speed of the pin} \\ \\ V1=V2\cdot0.847=4.043m/s,\text{ the speed of the ball} \\ \end{gathered}[/tex]
I already answered part a and b, need help with part c and d.
a.
b.
c.
[tex]\begin{gathered} r^2=400^2+300^2 \\ r^2=160000+90000 \\ r=\sqrt[]{250000} \\ r=\text{ }500\text{ N} \end{gathered}[/tex]d.
[tex]\begin{gathered} \tan x=\frac{400}{300} \\ x=\tan ^{-1}1.33333333333 \\ x=53.1301023473 \\ x\approx53^{\circ} \end{gathered}[/tex]Kim pushes horizontally on a block with a force of 800N moving it through a distance of 5.0 m across a level floor in a time of 20s. What is the power used by Kim? A ) 200W B) 80W C) 800W D) 2.0W
A ) 200W
Explanationpower is the amount of energy transferred or converted per unit time
[tex]\begin{gathered} Power=\frac{W}{t} \\ where\text{ P}is\text{ the power \lparen Watts\rparen} \\ W\text{ is the work done} \\ \text{ t is the time} \end{gathered}[/tex]also,• Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force
[tex]Work(W)=Force\text{ *distance}[/tex]therefore, replacing we have
[tex]\begin{gathered} Power=\frac{W}{t} \\ Power=\frac{Force*distance}{time} \end{gathered}[/tex]
hence,
Step 1
a)let
[tex]\begin{gathered} Force=800\text{ N} \\ distance=5.0\text{ m} \\ time\text{ = 20 s} \end{gathered}[/tex]b) now, replace in the formula:
[tex]\begin{gathered} Power=\frac{Forced\imaginaryI stance}{t\imaginaryI me} \\ Power=\frac{800\text{ N*5.0 m}}{20\text{ s}} \\ Power=200\text{ Watts} \end{gathered}[/tex]therefore, the answer is
A ) 200W
I hope this helps you
Given a ball moving in free fall between two heights. The ball moves from an initial height to some final height while still moving. If the initial gravitational potential energy of the ball is 5.00 J and the initial kinetic energy of the ball is 5.00 J then if the final gravitational potential energy of the ball is 2.00 J then what is the final kinetic energy of the ball? A) 2.00 JB) 5.00 JC) 8.00 JD) 106 JE) 10.0 j
ANSWER
C) 8.00 J
EXPLANATION
Given:
• The initial gravitational potential energy, 5.00 J
,• The initial kinetic energy, 5.00 J
,• The final gravitational potential energy, 2.00 J
Unknown:
• The final kinetic energy
By the law of conservation of energy, we know that the total mechanical energy must be conserved.
A ball moving in free fall only has gravitational potential energy and kinetic energy. The sum of the two must be constant for all heights,
[tex]PE_i+KE_i=PE_f+KE_f[/tex]Replace with the known values,
[tex]\begin{gathered} 5.00J+5.00J=2.00J+KE_f \\ 10.00J=2.00J+KE_f \end{gathered}[/tex]Solving for the final kinetic energy,
[tex]KE_f=10.00J-2.00J=8.00J[/tex]Hence, the final kinetic energy of the ball is 8.00 J
In a gravity escape system, an enclosed lifeboat on a large ship is deployed by letting it slide down a ramp and then continue in free fall to the water below. Suppose a 4900-kg lifeboat slides a distance of 5.0 m on a ramp that makes an angle of 600 with the vertical. How much work does gravity do on the boat?
The amount of work done, by gravity on the boat, given that the boat slides a distance of 5.0 m on the ramp is 120050 J
How do I determine the work done?We know already that work is defined as the product of force and distance moved in the direction of the force.
Work done (Wd) = force (F) × distance (d)
Wd = Fd
Since some level of angle projection is involved, the above formula can be written as:
Wd = FdCosθ
With the above formula, we can detertmine the work done. Details below
From the question given above, the following data were obtained:
Mass (m) = 4900 KgDistance (d) = 5 mAngle (θ) = 60°Acceleration due to gravity (g) = 9.81 m/s²Force (F) = mg = 4900 × 9.8 = 48020 NWorkdone (Wd) =?Wd = FdCosθ
Wd = 48020 × 5 × Cos 60
Wd = 48020 × 5 × 0.5
Wd = 120050 J
Thus, we can conclude that the work done is 120050 J
Learn more about workdone:
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When String A is played with String B, itcreates a 1.0 Hz beat frequency. WhenString A is played with String C, itcreates a 2.0 Hz beat frequency. If youplay String B and C together, what beatfrequency will they make? (There aretwo possible answers; give one.)(Unit = Hz)
Given:
Frequency produced by String A and B = 1.0 Hz.
Frequency produced by String A and C = 2.0 Hz.
Let's find the frequency produced by string B and C.
Apply the formula:
[tex]f_{21}=|f_2-f_1|[/tex]Now, for the frequency produced by A and B, we have:
[tex]\begin{gathered} f_{BA}=|f_B-f_A| \\ \\ 1.0=f_B-f_A \\ \end{gathered}[/tex]For the frequency produced by A and C, we have:
[tex]\begin{gathered} f_{CA}=|f_C-f_A| \\ \\ 2.0=f_C-f_A \end{gathered}[/tex]Now, we have the system of equations:
1.0 = fB - fA
2.0 = fC - fA
Let's solve both equations simultaneously.
Rewrite the first equation for fA:
fA = fB - 1.0
Substitute (fB - 1.0) for fA in the second equation:
2.0 = fC - (fB - 1.0)
2.0 = fC - fB + 1.0
2.0 - 1.0 = fC - fB
1.0 = fC - fB
Therefore, the beat frequency string B and string C will make when played together is 1.0 Hz.
• ANSWER:
1.0 Hz.
Two cars are headed in the same direction on the HWY. The trailing car is moving at14.6 m/s and has a mass of 1,230 kg. The lead car is moving at 13 m/s and has a massof 1,278 kg. The trailing car runs into the lead car and bumps it. Afterwards, the trailingcar has a velocity of 13.6 m/s.To determine if the collision is elastic, calculateΔKE = ΚΕ' – KE
The initial momentum of the system is,
[tex]p_i=m_1u_1+m_2u_2[/tex]The final momentum of the system is,
[tex]p_f=m_1v_1+m_2v_2[/tex]According to conservation of momentum,
[tex]p_i=p_f[/tex]Plug in the known values,
[tex]m_1u_{1_{}}+m_2u_2=m_1v_1+m_2v_2[/tex]Substitute the known values,
[tex]\begin{gathered} (1230kg)(14.6m/s)+(1278kg)(13m/s)=(1230kg)(13.6m/s)+(1278kg)v_2 \\ 17958\text{ kgm/s+}16614\text{ kgm/s=}16728\text{ kgm/s+}(1278kg)v_2 \\ v_2=\frac{34572\text{ kgm/s-}16728\text{ kgm/s}}{1278\text{ kg}} \\ =13.96\text{ m/s} \end{gathered}[/tex]The initial kinetic energy of the system is,
[tex]K=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_2u^{2^{}}_{2^{}}[/tex]The final kinetic energy of the system is,
[tex]K^{\prime}=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^{2^{}}_{2^{}}[/tex]The change in kinetic energy of the system is,
[tex]\Delta K=K^{\prime}-K[/tex]Plug in the known expressions,
[tex]\begin{gathered} \Delta K=\frac{1}{2}m_2v^2_2+\frac{1}{2}m_1v^2_1-\frac{1}{2}m_2u^2_2-\frac{1}{2}m_1u^2_1 \\ =\frac{1}{2}m_2(v^2_2-v^2_1)+\frac{1}{2}m_1(v^2_1-u^2_1) \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} \Delta K=\frac{1}{2}(1278kg)((13.96m/s)^2-(13m/s)^2)+ \\ \frac{1}{2}(1230kg)((13.6m/s)^2-(14.6m/s)^2) \\ =(639\text{ kg)(}194.88m^2s^{-2}-169m^2s^{-2})+(615\text{ kg)(}184.96m^2s^{-2}-213.16m^2s^{-2}) \\ =(639\text{ kg)(}25.88\text{ }m^2s^{-2})(\frac{1\text{ J}}{1\text{ kg}m^2s^{-2}})+(615\text{ kg)(}-28.2m^2s^{-2})(\frac{1\text{ J}}{1\text{ kg}m^2s^{-2}}) \\ =16537.32\text{ J-}17343\text{ J} \\ =-805.68\text{ J} \end{gathered}[/tex]Therefore, the change in kinetic energy of the system is -805.68 J.
Using the data provide in the following image could you help with this question If acceleration is supposed to be 9.8 m/s2 [down], determine the percentage error of your value. And State some possible causes of the error in your measured value. What techniques could be used to correct it?
It takes a worker 1,976 Joules of work to pull a cart 11 meters with a cable. If he is pulling at an angle of 65 degrees from up from the ground, what is the force of tension in the cable?
Answer: 425.056 N
Explanation:
We know,
W = Fscosα
⇒ 1976 = F × 11 × cos(65)
⇒ F = 425.056 N
2. In a race, if a runner starts and stops at the same position, what is theirdisplacement? *
Displacement is the shortest path covered by an object or body. As a runner starts and stops at the same positon, then the displacement of the runner equals to zero.
What process is illustrated in this diagram? A). The annihilation of a particle and its antiparticle to make two photonsB). The evaporation of a black hole C). The formation of an atom with a positive nucleus and a negative electron D). The creation of two photons from pure energy
Given a diagram of the creation of two photons.
In the diagram, an electron and a positron are combined to form two photons.
The electron is a matter. And positron is antimatter. In the diagram upon combining these two, pure energy is formed.
Thus, the diagram represents the annihilation of the particle(electron) and its antiparticle(positron) to form two protons.
Therefore, option A is the correct answer.
draw a Free-Body Diagram showing the forcesacting on the person.
The free body diagram of the elevator can be shown as,
Here, N is the normal force, W is the weight and F is the net force.
One force magnitude 72 acts on an object. Another force of magnitude on the abject is the magnitude of the resultant force on the object?
The magnitude, let us say, of force 1, F₁=72 units
The magnitude of the force that is acting at a right angle, F₂= 52 units
Let us assume that force 1 is acting along the positive x-axis and the second force is acting along the positive y-axis.
In the vector form, we can write,
[tex]\begin{gathered} \vec{F_1}=72\text{ }\hat{\text{i}} \\ \vec{F_2}=52\text{ }\hat{\text{J}} \end{gathered}[/tex]Thus the sum of these two forces will be,
[tex]\begin{gathered} \vec{F}=\vec{F_1}+\vec{F_2} \\ =72\text{ }\hat{\text{i}}+52\text{ }\hat{\text{j}} \end{gathered}[/tex]The magnitude of a vector
[tex]\vec{A}=\vec{B}+\vec{C}[/tex]is given by,
[tex]A=\sqrt[]{B^2+c^2}[/tex]Therefore the magnitude of vectr F is given by,
[tex]\begin{gathered} F=\sqrt[]{F^2_1+F^2_2} \\ \end{gathered}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} F=\sqrt[]{72^2+52^2} \\ =88.81\text{ units} \end{gathered}[/tex]Thus the magnitude of the resultant vector is 88.81 units
One closed organ pipe has a length of 1.65 meters. When a second pipe is played at the same time, a beat note with a frequency of 1.8 hertz is heard. By how much is the second pipe too long? Include units in your answer.
Given,
The length of one of the organ pipes, L₁=1.65 m
The frequency of the beat note, f=1.8 Hz
The speed of sound is given by, v=343 m/s
The beat frequency is given by,
[tex]\begin{gathered} f=|f_1-f_2| \\ =|\frac{v}{4L_1}-\frac{v}{4L_2}| \end{gathered}[/tex]Where f₁ is the frequency of the 1st pipe, f₂ is the frequency of the second pipe, and L₂ is the length of the second pipe.
On rearranging the above equation,
[tex]\begin{gathered} \frac{4f}{v}=|\frac{1}{L_1}-\frac{1}{L_2}| \\ \Rightarrow\frac{1}{L_2}=\frac{1}{L_1}-\frac{4f}{v} \\ =L_2=\frac{1}{\frac{1}{L_1}-\frac{4f}{v}} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} L_2=\frac{1}{\frac{1}{1.65}-\frac{4\times1.8}{343}} \\ =\frac{1}{0.61-0.02} \\ =1.7\text{ m} \end{gathered}[/tex]The difference in the length of the organ pipes is,
[tex]\begin{gathered} \Delta L=L_2-L_1 \\ =1.7-1.65 \\ =0.05\text{ m} \end{gathered}[/tex]Therefore the second pipe is longer than the 1st pipe by 0.05 m
The hours of daylight can vary greatly because of latitude.How would being near the equator shape the amount of sunlight per day in those countries? Would a La Niña event have a negative impact on wildlife?Why do you think some people have a difficult time believing that global warming is real?
The countries nearer to the equator always have an "average" daylength across all year, and they are "nearer" to the sun overall so the sunlight levels are slightly greater.
La niña event has always negative and positive impact on wildlife.
The fact that people have a difficult time believing global warming tends to be more sociological than factual.
A flashlight uses batteries that add up to 6.0 V and has a power output of 0.51 W.(a) How much current is drawn from the batteries?______ A(b) What is the effective resistance of the flashlight? _______Ω
a)
We know that the power is given by:
[tex]P=VI[/tex]where V is the voltage and I is the current. In this case the voltage is 6 V and the power is 0.51 W; plugging these values and solving for the current we have that:
[tex]\begin{gathered} 0.51=6I \\ I=\frac{0.51}{6} \\ I=0.085 \end{gathered}[/tex]Therefore, the current is 0.085 A.
b)
Using Ohm's law we have that:
[tex]\begin{gathered} R=\frac{6}{0.085} \\ R=70.59 \end{gathered}[/tex]Therefore, the resistance is 70.59 Ohms
A 55.9-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.621 and 0.284, respectively.What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?
The force is required to (a) just start the crate moving =340.196 N
(b) slide the crate across the dock at a constant speed =155.58 N
Given - mass m=55.9kg , μ=0.621 , μₙ=0.284
To find Horizontal pushing Force for just start the crate moving , static coefficient is applicable , while for other case kinetic coefficient is applicable.
concept used :-
F= μN ⇒μmg ⇔ For just start the crate moving
f=μₙN ⇒μₙmg ⇔For sliding the crate across the dock
Calculations :-
F= 0.621 x 55.9 x 9.8 =0.621 x 547.82
=340.196 N
f= 0.284 x 55.9 x 9.8 =0.284 x 547.82
=155.58 N
Result:-
Hence the force is required to (a) just start the crate moving =340.196 N
(b) slide the crate across the dock at a constant speed =155.58 N
To know about coefficients of static and kinetic friction :-https://brainly.com/question/17237604
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Give an example of each of the different types of energies: Kinetic; Potential; Thermal; Chemical; Nuclear; Sound; Electrical; Electromagnetic. Write the energy type and give an example of it.
Kinetic energy is the energy produced due to the motion of objects.
Example: A body moving with velocity.
Potential energy is the energy due to the position of objects.
Example: Any object placed at a certain height, h.
Thermal energy is the energy due to the temperature of the substance.
Example: Boiling water has thermal energy.
Chemical energy is the energy due to bonds formed between chemical compounds.
Example: Baking soda and vinegar when mixed form new chemical bonds that possess energy.
Nuclear energy is the energy due to the nucleus of atom.
Example: In nuclear fission and nuclear fusion energy is produced due to change in the nucleus.
Sound energy is the energy due to the vibration of objects.
Example: Tuning fork is one such example that produces sound.
Electrical energy: This energy is produced due to the movement of charged particles.
Example: Battery
Electromagnetic energy is the energy due to the emission and absorption of electromagnetic waves.
Example: Radio waves
I need help solving this problem. It’s really giving me a hard time. I need to know the formula and how to solve .
Take into account that the formula for the momentum is:
p = m*v
where,
v: velocity
m: mass
1) In this case, you have:
mA = 35 kg
vA = 5 m/s
The momentum is:
p = (mA)(vA) = (35 kg)(5 m/s) = 175 kg*m/s
2) If son B pushes on son A with a force of F = 40N, then, by the Newton third law, son A pushes on son B with the same force F = 40 N
3) To determine the acceleration of son B, use the following formula:
F = mB*a
where a is the acceleration. If you solve for a and repalce the values of other parameters, you obtain:
a = F/mB
a = 40 N/ 20 kg = 2 m/s^2
The temperature of outer space is about 3 K. What is this temperature on the Celsius scale? On the Fahrenheit scale? (Assume temperatures to be exact numbers.)(a) the Celsius scale °C(b) the Fahrenheit scale °F
Given the temperature:
3 K
Let's convert the temperature from Kelvin to degrees Celsius and Fahrenheit.
• (a). The Celsius scale:
To convert to celsius, we have:
[tex]\mathring{}C=K-273.15[/tex]Where K = 3, we have:
[tex]\mathring{}C=3-273.15=-270.15^oC[/tex]Therefore, the temperature on the Celsius scale is -270.15 °C.
• (b). , The Fahrenheit scale:
To convert from Kelvin to Fahrenheit, apply the formula:
[tex]^oF=1.8*(K-273.15)+32[/tex]Substitute 3 for K and solve:
[tex]\begin{gathered} ^oF=1.8*(3-273.15)+32 \\ \\ ^oF=1.8*(-270.15)+32 \\ \\ ^oF=−486.27+32 \\ \\ ^oF=−454.27 \end{gathered}[/tex]Therefore, the temperature in Fahrenheit is -454.27 °F.
ANSWER:
• (a). , -270.15 °C.
• (,b). , -454.27 °F.
How much spring potential energy (in Joules) can be held inside a spring with a spring constant of 335 N/m and a maximum elastic stretch of 23.7 cm?
The elastic potential energy of a spring is given by:
[tex]U=\frac{1}{2}kx^2[/tex]Then for this spring we have:
[tex]U=\frac{1}{2}(335)(0.237)^2=9.41[/tex]Therefore the potential energy is 9.41 J (rounded to two decimals)
A rope is used to pull a 1.39-kg bucket of water out of a deep well. In m/s^2, what is the acceleration of the bucket when the tension in the rope is 20.2 N?Answer: ________ m/s^2
Answer:
4.73 m/s²
Explanation:
The free body diagram for the bucket is
Then, the net force is equal to
Fnet = T - mg = ma
Where T is the tension of the rope, m is the mass, g is the gravity, and a is the acceleration. Solving for a, we get:
[tex]\begin{gathered} T-mg=ma \\ \\ a=\frac{T-mg}{m} \end{gathered}[/tex]Replacing T = 20.2 N, m = 1.39 kg, and g = 9.8 m/s², we get
[tex]\begin{gathered} a=\frac{20.2\text{ N - \lparen1.39 kg\rparen\lparen9.8 m/s}^2)}{1.39\text{ kg}} \\ \\ a=\frac{20.2\text{ N - 13.622 N}}{1.39\text{ kg}} \\ \\ a=\frac{6.578\text{ N}}{1.39\text{ kg}} \\ \\ a=4.73\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration is 4.73 m/s²
How can I find the missing current flow from the circuit?
Given:
Two resistances are connected in series.
A current of 0.4 A flows through the circuit.
A current of 0.2 A flows through a resistor.
To find The current through another resistor which is connected in series.
Explanation:
When two resistors are connected in series, the current flowing through them will be the same.
Final Answer: The current flow through the resistor of the circuit is 0.2 A
97) P = 2i + 2w for i
The given equation is:
p = 2i + 2w
To solve for i:
Step 1: Subtract 2w from both sides
p - 2w = 2i + 2w - 2w
p - 2w = 2i
Step 2: Divide both sides by 2
[tex]\begin{gathered} \frac{p-2w}{2}=\frac{2i}{2} \\ \frac{p-2w}{2}=\text{ i} \end{gathered}[/tex]This can then be re-written as:
[tex]i\text{ = }\frac{p-2w}{2}[/tex]