How much energy has 4×10^10m^3 of water collected in a reservoir at a hight of 100 m from the power house ?What kind of energy is that?

Answers

Answer 1

Answer:

PE = 3.92x10^16J

potential energy

Explanation:

PE = m*g*h

mass of water = 1000kg/m³

(4*10^10m³)*1000kg = 4*10^13kg

PE = (4*10^13kg)*(9.81m/s²)*(100m)

PE = 3.92x10^16J


Related Questions

What is the meaning of the word "force"?

Answers

Answer:

Explanation:

The word force has a precise meaning. A force is exerted on one object by another. The idea of a force is not limited to living things or non-living things. All objects (living and non-living) can apply a force on or to another object also all objects (living and non living) can be affected by forces.

Compared to the inertia of a 1-kilogram mass, the inertia of a 4-kilogram mass is​

Answers

Explanation:

The inertia of a 4 kg mass is four times as great as a 1 kg mass.

Question 5 of 10
A 15 N force is applied to an object, which then accelerates at 5.0 m/s2. What
is the mass of the object?
A. 10 kg
B. 20 kg
C. 75 kg
D. 3.0 kg

Answers

Hi there!

[tex]\large\boxed{D. \text{ } 3.0 kg}}[/tex]

Use the following equation to solve:

F = ma, where:

F = Force (N)

m = mass (kg)

a = acceleration (m/s²)

Thus:

15 = 5m

15/5 = m

mass = 3 kg

D is the correct answer

The search for black holes involves searching for The search for black holes involves searching for Group of answer choices single stars that emit large amounts of X-rays. a stellar core greater than 3 solar masses large spherical regions from which no light is detected. pulsars with periods less than one millisecond. pulsars that are orbited by planets.

Answers

Answer:

Large spherical regions from which no light is detected

Explanation:

A black hole is an object that has an extremely high density such that it possesses very powerful gravitational force that prevents the escape of all objects including light from it, and consumes nearby objects.

Due to the power of the gravitational force of a black hole, at the center, objects are infinitesimally compressed resulting in the inapplicability of the concept of space and time and the location is known as a singularity

Therefore, the search for black holes involves searching for large spherical regions from which no light is detected.

Can somebody help me understand this

Answers

I think it’s C. Marshmallow and toothpicks are used to show the composition of a water molecule

what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I

Answers

Answer: hello some part of your question is missing attached below is the missing detail

answer :

wf = M( v cos∅ )D / I

Explanation:

The Angular speed wf  of the system after collision in terms of the system parameters  and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3  + mD^2 ) wf

where ;  ( ML^2 / 3  + mD^2 ) = I ( Inertia )

In terms of system parameters and I

wf = M( v cos∅ )D / I

what are MA and VR of a lever?​

Answers

Explanation:

Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time

: Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2. Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?

Answers

Answer:

The angle is 16 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?

Surface charge density,   σ = 4 x 10^-9 C/m^2

charge, q = 10^-8 C

mass, m = 0.008 kg

Let the tension is the string is T and the angle is A.

[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree[/tex]

Clothes are flung to the sides of a round tub during the spin-cycle of a washing machine.If the tub's period of rotation is 0.2s and the tub's radius is 35cm,answer the related qestions below: (a)The frequency of roation (b)The circumference of rotation (c)The spinning of speed (d)The centripetal acceleration (e)The centripetal force of rotation if total mass rotated is 30kg (f)The power of washing machine​

Answers

Answer:

Explanation:

First I need to tell you that I used .20 s for the period of rotation instead of just .2, and I used 30.0 kg for the mass instead of just 30. The reason being that both those numbers as stated in the problem only have 1 significant digit and that's not generally enough to get the accuracy you're looking for. Adding a 0 to the ends of each of those numbers doesn't change the value of the numbers, only the number of sig fig's in each. Beginning with a:

a. [tex]f=\frac{1}{T}[/tex] so [tex]f=\frac{1}{.2}[/tex] and f = 5.0 Hz

b. The circumference is the distance around the outside of the washer's drum. We need to find that, but before we do, I'm going to state the radius in meters instead of cm. 35 cm = .35 m. Therefore,

C = 2(3.1415)(.35) so

C = d = 2.2 m

c. The speed of the washer is found in d = rt, where r is the rate and our velocity and d is the distance around the outside of the drum (circumference). Therefore,

2.2 = v(.20) so

v = 11 m/s

d. The centripetal acceleration has an equation

[tex]a_c=\frac{v^2}{r}[/tex] so

[tex]a_c=\frac{(11)^2}{.35}[/tex] and

[tex]a_c=\frac{121}{.35}[/tex] so

[tex]a_c=350\frac{m}{s^2}[/tex]

e. The centripetal force has an equation

[tex]F_c=\frac{mv^2}{r}[/tex] and

[tex]F_c=\frac{(30.0)(11)^2}{.35}[/tex] and

[tex]F_c=[/tex] 1.0 × 10⁴ N

f. The equation for Power is

[tex]P=\frac{W}{t}[/tex] where W is work and W = FΔx (force times displacement). Therefore,

[tex]P=\frac{(1.0*10^4)(2.2)}{.20}[/tex] so

P = 1.1 × 10⁵ Watts

Do the data for the first part of the experiment support or
refute the first hypothesis? Be sure to explain your
answer and include how the variables changed in the
first part of the experiment.
Mechanical equivalent of heat

Answers

Answer:

Sample Response: The data for the first part of the experiment support the first hypothesis. As the height of the cylinder increased, the temperature of the water increased. At a greater height, the cylinder has more gravitational potential energy. This gravitational potential energy was completely converted to thermal energy, which increased the temperature of the water.

Explanation:

this the sample response from ED mechanical equivalent of heat lab.

Answer:

Sample response:

Explanation:

The data for the first part of the experiment support the first hypothesis. As the height of the cylinder increased, the temperature of the water increased. At a greater height, the cylinder has more gravitational potential energy. This gravitational potential energy was completely converted to thermal energy, which increased the temperature of the water.


Sonny took 16 seconds to finish a 100-m race. What was
Sonny's average speed in the race

Answers

Speed = Distance/Time

=> Speed = 100m/16s

=> Speed = 6.25m/s

=> Speed = 6.25 ms^-1

What is the net force on a remote controlled airplane if a force of 57.6N is applied West and a force of 95.7N is applied South?

Answers

Answer:

112 N at 239°

Explanation:

Applying,

Pythagoras theorem,

a² = b²+c².................. Equation 1

Where a = net force

From the question,

Given: b = 57.6 N, c = 95.7 N

Substitute into equation 1

a² = 57.6²+95.7²

a² = 3317.76+9158.49

a² = 12476.25

a = √12476.25

a = 111.7 N

a ≈ 112 N

Then the direction is given as

∅ = tan⁻¹(c/b)........... Equation 2

∅ = tan⁻¹(95.7/57.6)

∅ = tan⁻¹(1.66)

∅ = 58.9°

∅ ≈ 59°

Since the angle is in the third quadrant,

∅' = 59+180

∅ = 239°

Hence the right answer is

112 N at 239°

The 75.0 kg hero of a movie is pulled upward at a constant velocity by a rope. What is the tension on the rope?

Answers

Answer:

750 N

Explanation:

the tension on the rope is the weight of the hero

The force that the left team pulls with is 1000 N. If the right team's total mass is 300 kg and they accelerate by 1.2 m/s2, what is the force of resistance on the right team

Answers

Answer:

the force of resistance on the right team is 360 N

Explanation:

Given;

force of the left team, = 1000 N

total mass of the right team, m = 300 kg

acceleration of the right team, a = 1.2 m/s²

The force of resistance of the right team is calculated as;

Force = mass x acceleration

Force, F = 300 x 1.2

Force = 360 N

Therefore, the force of resistance on the right team is 360 N

Help me and I'll make u Brainliest and follow u

Answers

26. D. crushing the sugar cube and dissolving it in water.

27. A. atom

28. B. molecule

29. B. plum pudding model of Joseph John Thomson

30. B. He used cathode ray tubes which showed that all atoms contain tiny negatively charged subatomic particles or electrons.

31. D. protons and neutrons are relatively heavier than electrons.

A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.
(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points

Answers

Answer:

a)   t = 20 s,  b)  x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

Explanation:

This is a kinematics exercise

a) in minimum time to stop,

           v = vo + at

           v = 0

           t = -v0 / a

we calculate

          t = -100 / (5.00)

          t = 20 s

b) Let's find the length you need to stop

          v² = vo² + 2 a x

          x = -v0 ^ 2 / 2a

          x = - 100² / 2 (-5.00)

          x = 1000 m

As the runway is only 800 m long, the plane cannot land at this distance.

A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?
a) x4
b) x207.36
c) x0.25
d) x36

Answers

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

4. An object is thrown from from the ground upward with an initial speed of 3.75 m/s. How long will the object be in the air before it lands on the ground?​

Answers

Answer:

Explanation:

There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.

Going with the physics version first, here's what we know:

a = -9.8 m/s/s

v₀ = 3.75 m/s

t = ??

That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:

v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:

v = v₀ +at and filling in:

0 = 3.75 + (-9.8)t and

-3.75 = -9.8t so

t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:

2(.38) = .76 seconds. Now onto the calculus way:

The position function is

[tex]s(t)=-4.9t^2+3.75t[/tex] The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:

v(t) = -9.8t + 3.75 and

0 = -9.8t + 3.75 and

-3.75 = -9.8t so

t = ,38 and multiply that by 2 to find the time the whole trip took:

2(.38) = .76 seconds.

How do the rocks in the mantle move?
A. They do not move.
B. They move in convection currents.
C. They flow like water.
D. They move on top of the crust.

Answers

Answer:

b. they move in convection currents.

Explanation:

i learned this in 4th grade

What is the current flowing through the circuit shown? (V= 110 V, R, = 200, R2 = 300, R3 = 10 0) (Ohm's law: V = IR)
A. 1.8 A
B. 20 A
C. 0.05 A
D. 0.55 A​

Answers

Answer:

A

Explanation:

The first thing you have to do is go back and list the resistances correctly. R1 = 20 R2 = 30 and R3 = 10.

Leave the units off if you can't make an omega.

The resistance of a series circuit (that's what this is) is r1 + r2 + r3 = 10 + 20 + 30 = 60 ohms

Now use ohms law.

R = 60 ohms

V = 110 volts.

I = V / R

I = 110/60

I = 1.833 to the nearest 1/10 = 1.8

Why do scientists use science

Answers

Answer:

Firstly, science helps our understanding of the world around us. Everything we know about the universe, from how trees reproduce to what an atom is made up of, is the result of scientific research and experiment.

Scientists use the scientific method to collect measurable, experimental evidence in an experiment related to a term (often in the form of an if/then statement), the results aiming to support or differ a theory. In other words, scientific method helps scientists get accurate, repeatable results.

how to answer thiss helppp​

Answers

is that all the info it has?

Answer:

Look at work

Explanation:

Your given the mass of planet M and radius as well as second mass.

Use universal gravitation law to find force of gravity

Fg= Gm1m2/r^2

Plug in givens

Fg= 553.6 N

Use newtons second law

Fg= ma

553.6N=5a

solve for a= 110.72m/s

Weight is 553.6N

What kind of waves are present during an earthquake?​

Answers

Body Waves - Primary (P) & Secondary (S) Waves.
Surface Waves - Rayleigh & Love Waves.

Answer:

There are four main types of earthquake waves:

P-waves S-waves (which are body waves), Rayleigh waves Love waves (which are surface waves).

HOPE IT HELPS!!

The most powerful empire between the 1500s and 1600s was the __________ Empire.
A.
Ottoman
B.
Mauryan
C.
Roman
D.
Persian

Answers

Answer:

A

Explanation:

Answer:

Ottoman

Explanation:

siri told me after I asked

Using diagram 1.1 and diagram 1.2, compare the number of turn of the coils, the pattern of the iron fillings and the angle of deflection of the ammeter pointer

Answers

Answer:

The number of turns in the second coil is more than the coil 1.

Explanation:

The magnetic field lines are the imaginary path on which an isolated north pole moves if it is free to do so.

The tangent at any point to the magnetic field line, gives the direction of magnetic field at that point.

More be the crowd ness of magnetic field lines more is the strength of magnetic field.

Here the crowd ness of magnetic field lines is more in figure 2 , so the magnetic filed in figure 2 is more than 1. It shows that the number of turns in the second coil is more than the 1 and also the current in the coil 2 is more than 1 .

Un auto que acelera a razón de 2 m/S^2 ; Si para cierto tramo se observa que logra triplicar su rapidez en 6 s. ¿Cuál fue su rapidez al final de estos 6 segundos?

Answers

Answer:

12m/s

Explanation:

Using the equation of  motion

v = u + at

Given that

u = 0m/s

a = 2m/s²

time t = 6s

Substitute the given values into the expression

v = 0 + 2(6)

v = 0+12

v = 12m/s

Hence the required velocity is 12m/s

what is acceleration

Answers

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

[tex]\sf What \: is \: acceleration? \\ \\ \sf The \: rate \: of \: change \: of \: velocity \: of \: an \\ \sf object \: with \: respect \: to \: time \\ \sf is \: known \: as \: acceleration. [/tex]

what do you mean by time?​

Answers

Time is the continued sequence of existence and events that occurs in an apparently irreversible succession from the past, through the present, into the future.

Find the transformation matrix that rotates a rectangular coordinate system through an angle of 60 about axes equal angels with original three coordinate axes

Answers

Answer:

  [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

Explanation:

To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system

          x ’= x cos (-60)

          y ’= y sin (-60)

we use

          cos 60 = cos (-60)

          sin 60 = - sin (-60)

we substitute

          x ’= x cos 60

          y ’= - y sin 60

the transformation system is

         [tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}cos 60&0\\0&-sin60\end{array}\right] \ \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]x '

the transformation matrix is

       [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

pahelp po ako.

1. Which of the following terms refers to energy in motion?
A. kinetic energy C. potential energy
B. light energy D. thermal energy

2. Which of the following is a type of energy gained due to its position?
A. kinetic energy C. gravitational potential energy
B. light energy D. thermal energy

3. How do you get the total mechanical energy of an object in a closed system?
A. Get the sum of the initial and final kinetic energy.
B. Get the sum of the initial and final potential energy.
C. Get the sum of the kinetic energy and the potential energy.
D. Get the difference between the kinetic energy and the potential energy.

4. Which of the following is the best set up to observe the conservation of mechanical energy?
A. A cart rolling down a wooden ramp.
B. A wooden block falling in a vacuum.
C. A volleyball falling from a tall building.
D. a paper plane being launched from a tower.

5. A big rock was dropped from the third floor of a building. What happens to the kinetic energy of the ball as it falls?
A. The rock’s kinetic energy increases.
B. The rock’s kinetic energy decreases.
C. The rock’s kinetic energy remains the same.
D. The rock’s kinetic energy cannot be identified.

6. A heavy metallic ball was rolled up on a frictionless ramp. What happens to the gravitational potential energy of the ball as it rolls up the ramp?
A. The ball’s gravitational potential energy increases.
B. The ball’s gravitational potential energy decreases.
C. The ball’s gravitational potential energy remains the same.
D. The ball’s gravitational potential energy cannot be identified.


7. A 2kg ball is elevated up a building, how much is its potential energy 10m from the ground?
A. 5J C. 98J
B. 20J D. 196J

For items 8 to 10, refer to the illustration below. A ball is dropped at point W and travels along a frictionless path to reach point Z.
8. At which point does the ball has greatest gravitational potential energy?
A. W C. Y
B. X D. Z

9. At which point does the ball has greatest kinetic energy?
A. W C. Y
B. X D. Z

10. What happens to the total mechanical energy of the ball as it travels from point W to point Z?
A. The total mechanical energy increases.
B. The total mechanical energy decreases.
C. The total mechanical energy remains the same.
D. The total mechanical energy cannot be determined.

For items 11 to 12, refer to the illustration below. A pendulum swings back and forth from its highest position to its lowest position and vice versa.

11. As a pendulum swings from its highest to lowest position, what happens to its gravitational potential energy and kinetic energy?
A. Both its GPE and KE increases.
B. Both its GPE and KE decreases.
C. Its GPE decreases but KE increases.
D. Its GPE increases but KE decreases.

12. What happens to the total mechanical energy of the bob as it swings back and forth if only force acting on it is the gravitational force?
A. The total mechanical energy increases.
B. The total mechanical energy decreases.
C. The total mechanical energy remains the same.
D. The total mechanical energy cannot be determined.
Four ball of masses 0.5 kg, 1 kg, 2 kg, and 5 kg were dropped simultaneously on top of a building with the height of 20 m. Ignoring the air resistance, the four balls reach the ground at the same time.
13. Which of the following balls has the greatest gravitational potential energy on top of the building?
A. The ball with the mass of 5 kg.
B. The ball with the mass of 2 kg.
C. The ball with the mass of 0.5 kg.
D. All the balls have the same potential energy.
14. Which of the following balls has the least kinetic energy on top of the building?
A. The ball with the mass of 5kg.
B. The ball with the mass of 1kg.
C. The ball with the mass of 0.5kg.
D. All the balls have the same kinetic energy.

15. Which of the following balls has the greatest total mechanical energy?
A. The ball with the mass of 2kg.
B. The ball with the mass of 1kg.
C. The ball with the mass of 0.5kg.
D. All the balls have the same total mechanical energy

Answers

D bc thats sound like the only resonable answer
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