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trasher [3.6K]
3 years ago
13

A basesball is dropped from 100 meters above the surface of the earth. The baseball falls to the ground. What would happen if th

is baseball was dropped from 100 meters above the surface of the moon?
Physics
2 answers:
Readme [11.4K]3 years ago
6 0

Answer:

It would fall toward the Moon's surface

Explanation:

As you already know, a ball that is thrown 100 meters above the surface of the earth, would fall on the surface of the earth, due to the gravity of our planet, which attracts any object around it to its surface (and that is what we keep them attached to the planet's floor). The same would happen if a ball were launched 100 meters above the surface of the moon, but the ball would fall on the surface of the moon, for the same reason that is the gravity of the moon. As the Moon is much smaller than the Earth, its force of gravity is also less, therefore the bodies are attracted with less force to the lunar surface, even so the ball would fall on the surface of the moon because the distance between the ball and the earth is much greater than the distance between the moon and the ball.

Serjik [45]3 years ago
4 0

Given that, A basesball is dropped from 100 meters above the surface of the earth. If the same baseball was dropped from 100 meters above the surface of the moon, it will take more time to hit the ground as compare to the ball dropped on earth. This is moon's gravity is one-sixth of that of earth. Object falling on earth possess more force of attraction. So it will reach the earth in lesser time. At moon the force of attraction is low compare to that of earth. Object will take more time to reach the surface.

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A theory is a hyothesis that has been varified by multiple investigations. <br> True or false
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The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
3 years ago
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