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ANTONII [103]
3 years ago
12

In a balloon adventure, you reach 10 km in height. Find the percentage variation in the weight of the balloon-crew system with r

espect to sea level.
Physics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

0.31%

Explanation:

The acceleration due to gravity at sea level is:

g = GM / R²

where G is the gravitational constant, M is the mass of Earth, and R is the radius of the Earth.

The acceleration due to gravity a height h above sea level is:

g' = GM / (R+h)²

The percent variation is:

(g − g') / g

= 1 − g'/g

= 1 − [GM / (R+h)²] / [GM / R²]

= 1 − R² / (R+h)²

= 1 − (R / (R+h))²

The average radius of the Earth is 6370 km.  The height of the balloon is 10 km.

1 − (6370 / (6370 + 10))²

0.0031

0.31%

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a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s
Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
3 years ago
After a massive-star supernova, what is left behind?.
scoray [572]

Answer: A <u>Nebula </u>is left behind. A spectacular explosion in which a star ejects most of its mass in a violently expanding cloud of debris.

Hope this helps!

8 0
3 years ago
Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th
Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion S=ut+\frac{1}{2}at^2 where s is distance a is acceleration and u is initial velocity

Using this equation for first plane 1200=0\times 30+\frac{1}{2}a30^2

a=2.67\frac{m}{sec^2}

As the acceleration is same for both the plane so a for second plane will be 2.67 \frac{m}{sec^2}

The another equation of motion is v^2=u^2+2as using this equation for second plane 40^2=0+2\times 2.67\times s

s = 300 m

5 0
3 years ago
Why does matter increase in volume when it heats up?
ArbitrLikvidat [17]

Answer:All three states of matter (solid, liquid and gas) expand when heated. ... Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.

Explanation:well I tried lol she just copied and pasted faster than I could

8 0
3 years ago
A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8
Arlecino [84]

Answer:

0.0025116weber/m²

Explanation:

Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).

Mathematically;

B = ¶/A

¶ = BA

Given B = 0.23Tesla which is the magnitude of the magnetic field

Dimension of the rectangular loop = 7.8 cm by 14 cm

Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm

= 109.2cm²

Converting this value to m²

Area of the loop = 109.2 × 10^-4

Area of the loop = 0.01092m²

Magneto flux = 0.23×0.01092

Magnetic flux = 0.0025116weber/m²

3 0
3 years ago
Read 2 more answers
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