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irina [24]
2 years ago
13

You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __

_________.
Physics
1 answer:
SVETLANKA909090 [29]2 years ago
5 0

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy {\rm GPE} of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

You might be interested in
The earth’s orbit around the sun is elliptical. What does this word mean?.
omeli [17]

Answer:

An elliptical orbit is the revolving of one object around another in an oval-shaped path called an ellipse. The planets in the solar system orbit the sun in elliptical orbits. Many satellites orbit the Earth in elliptical orbits as does the moon.

3 0
2 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is
sergiy2304 [10]

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

3 0
3 years ago
Please someone help, I’m very confused and it’s due soon, thanks
Anit [1.1K]

Answer:

  1. 1 s
  2. 19.6 m
  3. 2 s
  4. 0.8 m/s^2
  5. 28 m/s
  6. 79 m/s
  7. 0.37 s
  8. 26 m/s
  9. 242 m/s
  10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0
3 years ago
Someone stuck a piece of chewing gum on the outer edge of a blade on a ceiling fan. The radius (r) of where the gum is from the
maksim [4K]

Answer: why would u stick a gum there

Explanation:

3 0
3 years ago
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