Answer:
Magnitude of the force between the charges is F = 1.92×10^20N
Explanation:
Given the magnitude of force according to coulombs law
F =K[(q1*q2)/r2]
Where q1 and q2 are the charges
r is the distance between the charges
K is the coulombs constant
Substituting the given values, we have;
F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²
F = 43.1×10^19/2.25
F = 19.16×10^19N
F = 1.92×10^20N
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
Answer:
The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.
Explanation:
Let
denote the position vector of the ball hit by player A. Then this vector has components

where
is the magnitude of the acceleration due to gravity. Use the vertical component
to find the time at which ball A reaches the ground:

The horizontal position of the ball after 0.49 seconds is

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector
of the ball hit by player B has

Again, we solve for the time it takes the ball to reach the ground:

After this time, we expect a horizontal displacement of 12 meters, so that
satisfies


This sounds like thermal expansion.