There are a pair of forces ... two of them ... between you and the computer.
You're attracted to the computer, and the computer is attracted to you.
The forces have equal strength.
Force of gravity = (6.67 x 10⁻¹¹ N-m²/kg²) M₁· M₂ / (distance)²
= (6.67 x 10⁻¹¹ N-m²/kg²) (80 kg) (4 kg) / (0.25 m)²
= (6.67 x 10⁻¹¹ x 80 x 4 N-m²) / (0.125 m²)
= 1.71 x 10⁻⁷ Newton
= about 0.000 000 615 ounce .
THAT's why you don't feel especially attracted to your computer.
C, Land heats and cools faster than water!
Answer: the correct option is D (17m).
Explanation: The farthest distance at which a typical "nearsighted" frog can see clearly in air is 17m.
At its maximum height h, the football has zero vertical velocity, so if it was kicked with initial upward speed v, then
0² - v² = -2gh
Solve this for v :
v² = 2gh
v = √(2gh)
The height y of the football t seconds after being kicked is
y = vt - 1/2 gt²
Substitute v = √(2gh), replace y = h, and solve for h when t = 3.8 s :
h = √(2gh) t - 1/2 gt²
h = √(2gh) (3.8 s) - 1/2 g (3.8 s)²
h ≈ (16.8233 √m) √h - 70.756 m
(By √m, I mean "square root meters"; on its own this quantity doesn't make much physical sense, but we need this to be consistent with √h. h is measured in meters, so √h is measured in √m, too.)
h - (16.8233 √m) √h + 70.756 m = 0
Use the quadratic formula to solve for √h :
√h = ((16.8233 √m) ± √((16.8233 √m)² - 4 (70.756 m))) / 2
Both the positive and negative square roots result in the same solution,
√h ≈ 8.411 √m
Take the square of both sides to solve for h itself:
(√h)² ≈ (8.411 √m)²
⇒ h ≈ 70.756 m ≈ 71 m