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LUCKY_DIMON [66]
2 years ago
9

The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m3. Convert this volume to liters

and express the result in standard exponential notation.
Physics
1 answer:
Gwar [14]2 years ago
3 0

Answer:

3,666,500\ m^3 = 3.666\times 10^9\ Liters

Explanation:

Given that

Volume

V=3,666,500\ m^3

As we know that

1\ m^3 = 1000\ Liters

1\ m^3 = 10^3\ Liters

So

3,666,500\ m^3 = 3,666,500\times 10^3\ Liters

3,666,500\ m^3 = 3,666,5\times 10^5\ Liters

3,666,500\ m^3 = 3.666\times 10^9\ Liters

So we cab say that 3,666,500\ m^3 is equal to 3.666\times 10^9\ Liters.

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Describe why we cannot see details on the surface of Mars.​
bearhunter [10]

Answer: atmosphere

Explanation:

We cannot see the surface of Mars because of its thick atmosphere and of the iron dust that kicks up

4 0
3 years ago
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This is due by 11:59 PM tonight.
svetoff [14.1K]

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

6 0
3 years ago
Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
3 years ago
The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This
erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

g = \frac{GM}{(d-R_{CM})^2}

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

R_{CM} = Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

g = \frac{GM}{(d-R_{CM})^2}

g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}

g = 3.4*10^{-5}m/s^2

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

\omega = \frac{2\pi}{T}

At the same time we have that centripetal acceleration is given as

a_c = \omega^2 r

Replacing

a_c = (\frac{2\pi}{T})^2 r

a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)

a_c = 3.34*10^{-5}m/s^2

3 0
3 years ago
Find the magnitude of the sum<br> of these two vectors:<br><br> 101 m<br> 60.0 °<br> 85.0 m
attashe74 [19]

Answer: 161.3

I have a acellus too and got this question correct, so I hope this helps y’all out

8 0
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