Answer:
<em>Earth's gravity pulls air as close to the surface as possible. ... As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense than air nearer to sea level. This is what meteorologists and mountaineers mean by "thin air." Thin air exerts less pressure than air at a lower altitude.</em>
The correct answer is electromagnetic. Lights travel as an electromagnetic wave. In fact, light is an electromagnetic radiation which has a center of the electromagnetic spectrum. Light is visible to the naked eye, and is responsible for an organism's sense of sight.
Answer:
15 cm
Explanation:
Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:
Gaya 1 (F₁) = 225 N
Jarak terpisah 1 (d) = 5 cm
Gaya 2 (F₂) = 25 N
Jarak terpisah 2 (d₂) =?
Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:
F = Kq₁q₂ / d²
Perbanyak silang
Fd² = Kq₁q₂
Menjaga Kq₁q₂ konstan, kita memiliki:
F₁d₁² = F₂d₂²
Dengan rumus di atas maka diperoleh jarak sebagai berikut:
Gaya 1 (F₁) = 225 N
Jarak terpisah 1 (d) = 5 cm
Gaya 2 (F₂) = 25 N
Jarak terpisah 2 (d₂) =?
F₁d₁² = F₂d₂²
225 × 5² = 25 × d₂²
225 × 25 = 25 × d₂²
5625 = 25 × d₂²
Bagilah kedua sisinya dengan 25
d₂² = 5625/25
d₂² = 225
Hitung akar kuadrat dari kedua sisi
d₂ = √225
d₂ = 15 cm
Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N
hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
