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seropon [69]
4 years ago
8

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa

ce camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.
Physics
1 answer:
Ludmilka [50]4 years ago
7 0

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

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scoundrel [369]
I would say your answer is B- Some of the chemical energy from the batteries is converted into heat energy. 
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4 years ago
Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current
Alchen [17]

Answer:

M=0.0247H

Explanation:

Given data

V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H

4 0
3 years ago
Which one of the following energy conversion devices has the lowest energy efficiency?
Anna35 [415]

Answer:

A) incandescent ligth bulb, its efficiency is about 10%

Explanation:

The incandescent bulb, that is, the well-known focus with its warm light, was one of the most useful inventions of the 19th century although its use is currently considered very inefficient. These lamps waste between 80 and 90 percent of the total electricity they consume by turning it into heat. The metal filament thus heated and which is the central part of the bulb, only converts the remaining energy into light. Its service life ranges from 750 to 1,000 hours.

This is why they are used in ovens for food preparation, because of the large amount of heat they generate.

The steam boiler in a power plant depends on the fuel that it is using, but a coal-fired power plant with modern technology its efficiency is about 40%

Electric motor are around 85-92%

In order to better understand the concept of efficiency it is as if we pay 100 dollars of gasoline for our weekly use, but of that 100 dollars the car only uses 10 dollars to do that activity the rest of the money the 90 dollars were lost because of the inefficiencies of the vehicle.

3 0
3 years ago
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0
3 years ago
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Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

\lambda = \frac{v}{f}

Where,

\lambda = Wavelength

f = Frequency

v = Velocity

Our values are given as,

f = 2.8*10^3Hz

v = 340m/s \rightarrow Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

\lambda = \frac{340}{2.8*10^3}

\lambda = 0.1214m

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0
3 years ago
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