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astraxan [27]
2 years ago
13

The international space station (ISS) orbits the earth with a velocity 7.6 km/s. How much energy is required to lift a payload o

f 2500 kg from the surface of the earth to the ISS
Physics
1 answer:
Dmitry [639]2 years ago
4 0

Answer:

7.22×10¹⁰ J

Explanation:

From the question given above, the following data were obtained:

Velocity (v) = 7.6 Km/s

Mass (m) = 2500 kg

Energy (E) =?

Next, we shall convert 7.6 Km/s to m/s. This can be obtained as follow:

1 km/s = 10³ m/s

Therefore,

7.6 Km/s = 7.6 Km/s × 10³ m/s / 1 km/s

7.6 Km/s = 7.6×10³ m/s

Finally, we shall determine the energy required as follow:

Velocity (v) = 7.6×10³ m/s

Mass (m) = 2500 kg

Energy (E) =?

E = ½mv²

E = ½ × 2500 × (7.6×10³)²

E = 1250 × 57760000

E = 7.22×10¹⁰ J

Thus, 7.22×10¹⁰ J of energy is required.

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Two objects, C & D, have the same momentum. Object C has ½ the mass of object D. Find the value of the ratio of velocity C t
Savatey [412]
These are two questions and two answers.

Part 1. Fin the value of the ration of velocity C to velocity D.


Answer: 2

Explanation:

1) Formula: momentum = mass * velocity

2) momentum C = mass C * velocity C

3) momentum D = mass D * velocity D.

4) C and D have the same momentum =>

mass C * velocity C = mass D * velocity D

5) mass C = (1/2) mass D => mass C / mass C = 1/2

6) use in the equation stated in the point 4)

velocit C / velocity D = mass D / mass C

using the equation stated in point 5:

mass D / mass C = 1 / [ mass C / mass D] = 1 / [1/2] = 2

=>

7) velocity C / velocity D = mass D / mass C = 2

Part 2: <span>ratio of kinetic energy C to kinetic energy D.
</span>
Answer: 2

Explanation:

1) formula: kinetic energy KE = (1/2) mass * (velocity)^2

2) KE C = (1/2) mass C * (velocity C)^2

3) KE D = (1/2) mass D * (velocity D)^2

4) KE C / KE D =

(1/2) mass C * (velocity C)^2        mass C        (velocity C)^2
--------------------------------------- = --------------- * ---------------------- = (1/2) * (2)^2
(1/2) mass D *( velocity D)^2        mass D        v(velocity D)^2

= 4 / 2 = 2
3 0
3 years ago
To perform a drug lookup to ensure that the new compound has been added to the computer system properly , select in the toolbar
lutik1710 [3]

To perform a drug lookup to ensure that the new compound has been added to the computer system properly we have to select the new from the toolbar at the top of the screen

<h3>What is a computer system?</h3>

A computer system is a collection of computers, related hardware, and related software. The central processing unit (CPU), memory, input/output, and storage devices are the four main components of a computer system. To produce the desired result, all of these parts operate in concert as a single unit.

Selecting the new from the toolbar at the top of the screen will allow us to run a drug lookup to make sure the new compound has been properly added to the computer system.

Therefore the correct answer is the option C

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3 0
1 year ago
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

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What is the volume of a storage tank which will hold 3200kg of petrol?
otez555 [7]

Answer:

The volume of 3200 kg of petrol is 4 m^3.

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3 years ago
S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(
Ket [755]

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.       

V = kq/r

Where V represents electric potential, K is coulomb constant, q  is Charge and r is distance between any  two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V  = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

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3 0
1 year ago
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