A spinning satellite begins to unfold two solar panels as shown. As the
panels extend from the satellite, what is the result?

A Spinning Satellite Begins To Unfold Two Solar Panels As Shown. As Thepanels Extend From The Satellite,

Answers

Answer 1

Its angular momentum of a satellite will drop even as satellite panel stretch because of the reduction in angular velocity brought on by the drag effect.

Why then do stars spin more quickly during their demise?

A portion of the star's mass is blown off during the supernovae that come before the creation of black holes, taking some of the star's total angular momentum with it.The leftover material sinks into the star's core while continuing to spin rapidly.

What is the name of the spinning effect?

Magnus effect: When there is relative movement between both the spinning item and the fluid, a sideways force is generated on a rotating cylindrical or spherical object immersed in the fluid (liquid or gas).In honor of the German chemist and physicist H.G.

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Related Questions

how do you use the coefficient to calculate the number of atoms in each molecule?​

Answers

wait is there supposed to be a picture here?

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:

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?

Answers

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

Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. What is the swimmers average speed and average velocity?

Answers

Answer:

The average speed is 1 m/s

The average velocity is 0

Explanation:

Given;

length of the pool, L = 50 m

time taken for the motion, t = 100 s

The total distance = 50 m + 50 m

The total distance = 100 m

The average speed = total distance / total time

                                  = 100 / 100

                                  = 1 m/s

The average velocity = change in displacement / change in time

change in displacement = 50 m - 50 m = 0

The average velocity = 0 / 100

The average velocity = 0

A body is dropped from a height H. In how much time will it reach the ground?

Answers

Answer:

[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]

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​

Answers

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

Trong máy phát điện xoay chiều ba pha khi tổng điện áp tức thời của cuộn 1,2 là e1+e2=120V thì điện áp tức thời của cuộn 3 là

Answers

Answer:

I just noticd i dont speak this launguage

Explanation:

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.

Answers

Answer:

Explanation:

From the given information:

The initial PE [tex](PE)_i[/tex] = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = [tex]P.E_f -P.E_i[/tex]

ΔP.E = 0 - [tex](PE)_i[/tex]

ΔP.E = [tex]-P.E_i[/tex]

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]

[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]

this can be re-written as:

[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]

[tex]\Delta U =(0.70) (490.5)[/tex]

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?

Answers

Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field

Explanation:

While interpreting the data in NMR, the positions of signals are studied.

The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.

The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.

So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]

Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons

So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]

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)​

Answers

I= 0.39 A

OPTION B is the correct answer.

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

Answers

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]

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

Answers

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

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

Answers

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

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.

Answers

Answer

I hope this help....

Explanation:

Answer:

Hope this helps

Explanation:

please help very easy 5th grade work giving brainliest

Answers

Answer:

the answer is option B because opposit sides of the magnets attract each other

Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms​

Answers

Answer:

Sorry I don't know the answer

reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

Answers

Answer: θ would equal approximately 28.7°

This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.

Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°

Now if we multiply the range by 2, we get:

2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:

2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ

Thus, θ = 28.67780425

It's been awhile since I did this; though I hope it helped!

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

Answers

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

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. How much of the sample is left after 1.98 x 10^4 seconds?

A) 8.03 x 10^16 nuclei

B) 4.01 x 10^16 nuclei

C) 2.02 x 10^16 nuclei

D) 1.61 x 10^17 nuclei

Answers

OPTION C is the correct answer.

The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.

What is half-life?

The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.

The isotope decay of an atom is given by the equation:

ln [A] = -kt + ln [A]₀

The rate constant, k is:

k = ln 2 / Half life

k = ln 2 / 4.96 x 10³

k = 1.40 × 10⁻⁴ s⁻¹

t = 1.98 x 10⁴

[A]₀ = 3.21 x 10¹⁷

ln [A] = -1.40 × 10⁻⁴  ×  1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538

[A] = 2.02 x 10¹⁶ nuclei

Thus the correct option is C.

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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).

Answers

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

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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

Lighting is the movement of?

Answers

Explanation:

Movement:refers to the changing in the lights whether it be a change in intensity, color or direction of origin.

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?

Answers

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

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What is the energy equivalent of an object with a mass of 1.05g?​

Answers

Answer:

The equivalent energy of an object given its mass is calculated through the equation,

                             E = mc²

where c is the speed of light (3 x 10^8 m/s)

Substituting the known values,

                            E = (1.05 g/ 1000) (3 x 10^8 m/s)²

                               E = 9.45x10^13 J

Explanation:

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.

Answers

Answer:

66,7 seconds

Explanation:

the formula for height/distance is : S=v.t

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.

Answers

Answer:

564

Explanation:

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

Answers

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 recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet

Answers

Answer:

We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.

Explanation:

5 ways our governments can confront climate change

PROTECT AND RESTORE KEY ECOSYSTEMS

SUPPORT SMALL AGRICULTURAL PRODUCERS

PROMOTE GREEN ENERGY

COMBAT SHORT-LIVED CLIMATE POLLUTANTS

BET ON ADAPTATION, NOT JUST MITIGATION

What happens in the gray zone between solid and liquid?-,-​

Answers

The gray zone transition is very crucial which includes the inter molecular forces acting on the molecules and each atoms which makes the change in state from hot to cold and cold to hot. and for it to be liquid to solid or solid to liquid the transition needs to cross the gray zone.

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An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat

per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω

of the heater wire? (Note: 1 cal = 4.186 J)

Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43

Answers

Answer:

1 cal/s =4.184w

p=50 cal/s =2093w

v=12v

P = V*I

I =P/V

I = 17.43 A

P =1²*R

R = P/I²

R = 0.68

FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN

Answers

Clouds? I am not sure of your options!

From the water whole water cycle starts again.

Most possibly water should be the answer.

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

Answers

OPTION C is the correct answer.

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