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mario62 [17]
3 years ago
5

An office worker has his lunch on the 53rd floor of an office building. In which case is the gravitational potential energy of t

he lunch-Earth system greatest?
Physics
1 answer:
Ann [662]3 years ago
4 0

Options

a. The worker carried the lunch up to the 53rd floor in the elevator.

b. The worker got a ride in a helicopter to the top of the 68-floor building and then carried the lunch down in the elevator to the 53rd floor?

c. The Worker carried it up the stairs to the 53rd floor.

d. Gravitational potential energy does not depend on the path taken.

Answer:

The gravitational Energy would remain the same in all cases.

Explanation:

Gravitational Energy is a conservative energy and its potential energy comes from gravitational actions.

An instance is when someone lifts a weight from the floor to a certain height and then dropped the weight back to the floor; the work done by the field will not be altered at all; it'll be the same throughout.

If otherwise, it'll be against thermodynamics 2nd law of physics because the object would have remaining energy after returning to the starting point,

For this reason, the gravitational potential energy only depends from the height, and is independent from the path taken to reach there.

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At 20 C, a steel rod of length 40.000 cm and a brass rod
balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

<u>Step 1:</u> <u>identify the given parameters and standard parameters</u>

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

                     stress in the rod =?

 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

--------------------------------------------------------------------------------------------

<u>Step 2:</u> <u>calculate the stress in each rod</u>

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

<u>Step 3:</u> <u>calculate the new length of each rod</u>

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0
3 years ago
Which one of the following is not a fundamental physical quantity??! A. Temperature b. Current c. Area d. Mass​
Hatshy [7]

Answer:

Out of this, Area is not a fundamental physical quantity.

3 0
3 years ago
Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in fal
mote1985 [20]

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 m^{3}

height through which it falls, h is 150 m

mass of 1 m^{3} of water is 1000 kg

⇒mass of 1250 m^{3} of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 \frac{m}{sec^{2} }

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

        = 1250000×9.8×150 J

        = 1837500000 J

Electrical Energy = \frac{3}{4}(KE)

                            = \frac{3}{4}×1837500000

                            = <u>1378125000 J per second</u>

8 0
3 years ago
A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat
liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

PV=nRT

Here , n is no of moles , n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}

Putting initial and final values and dividing them :

\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}

\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.

Hence , this is the required solution.

7 0
3 years ago
If you wanted to produce more power,what could you do to maximize power
PSYCHO15rus [73]

In physics, power is defined as energy per unit time.  You will also hear it described as work per unit time.  The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time).  This is expressed as a fraction as J/s.  If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).

6 0
3 years ago
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