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4 years ago

7

increase of green house gases is warming the oceans and melting sea ice at the poles. This causes changes in climate. Name three

earth spheres which interact in this process. Justify your answer.

1 answer:

Ahat [919]4 years ago

4 0

core, surface atmosphere

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2. Suppose your car has a maximum braking acceleration of -5 m/s2. Calculate the stopping distance for an initial speed of 25 m/

Hitman42 [59]

Answer:

s=62.5m

Explanation:

Use the equation v²=u²+2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.

0²=25²+2(-5)s

10s=625

s=62.5m

3 0

3 years ago

A weightlifter lifts a 1400 N barbell 2.5 meters. Calculate the work done during the lift

Andre45 [30]

Answer:

Force = 1400N

Displacement = 2.5 m

work done = F×D

=1400×2.5

= 3500 joule

8 0

3 years ago

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What is substances ?????<br>

Shkiper50 [21]

<u>Substance</u><u>:</u><u>-</u>

A substance is matter which has a specific composition and specific properties.

7 0

3 years ago

A crane moves a 250 kg scoreboard from the ground to the height of 100 m. What is the work done on the scoreboard?

Mazyrski [523]

If you’re doing potential and kinetic energy then the answer is potential.

7 0

2 years ago

Read 2 more answers

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

KiRa [710]

Answer:

(a) Determine the astronaut's orbital speed

The velocity of the astronaut is 1436m/s.

(b) Determine the period of the orbit.

The period of the orbit is 10413 seconds.

Explanation:

<em>(a) Determine the astronaut's orbital speed</em>

The orbital speed of astronaut can be found by means of the Universal law of gravity:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Then, replacing Newton's second law in equation 3 it is gotten:

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a is the centripetal acceleration since the astronaut describes a circular motion around the Moon:

$a = \frac{v^{2}}{r}$ (3)

Replacing equation 3 in equation 2 it is gotten:

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (4)

Where v is the orbital speed, G is the gravitational constant, M is the mass of the Moon, and r is the orbital radius.

Notice that the orbital radius will be given by the sum of the radius of the Moon and the height of the astronaut above the surface.

But it is necessary to express the height of the astronaut above the surface in units of meters before it can be used.

$r = 680km \cdot \frac{1000m}{1km}$ ⇒ $680000m$

$r = 1.70x10^{6}m+680000m$

$r = 2380000m$

However it is neccesary to find the mass of Moon in order to use equation 4.

Then Newton's second law ( $F = ma$ ) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

Then, M will be isolated

$M = \frac{r^{2}a}{G}$ (5)

Where r is the orbital radius of the astronaut and a is the acceleration due to gravity.

$M = \frac{(2380000m)^{2}(0.867 m/s^{2})}{6.67x10^{-11}N.m^{2}/kg^{2}}$ (5)

$M = 7.36x10^{22}Kg$

$v = \sqrt{\frac{(6.67x10^{-11}N.m^{2}/kg^{2})(7.36x10^{22}Kg)}{2380000m}}$

$v = 1436m/s$

Hence, the velocity of the astronaut is 1436m/s.

<em>(b) Determine the period of the orbit. </em>

The period of the orbit can be determined by the next equation:

$v = \frac{2\pi r}{T}$ (6)

Where v is the orbital velocity, r is the orbital radius and T is the period of the orbit.

Then, T can be isolated from equation 6.

$T = \frac{2\pi r}{v}$ (7)

$T = \frac{2\pi (2380000m)}{1436m/s}$

$T = 10413s$

Hence, the period of the orbit is 10413 seconds.

6 0

3 years ago