Answer:
E_total = 1.30 10¹⁰ C / m²
Explanation:
The intensity of the electric field is
E = k q / r²
on a positive charge proof
The total electric field at the midpoint is
as q₁= 6 10⁻⁶ C the field is outgoing to the right
for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore
E_total = E₁ + E₂
E_total = k q₁ / r₁² + k q₂ / r₂²
r₁ = r₂ = r = 4 10⁻² m
E_total = k/r² (q₁ + q₂)
we calculate
E_total = 9 10⁹ / (4 10⁻²)² (6.0 10⁻⁶ +3.0 10⁻⁶)
E_total = 1.30 10¹⁰ C / m²
Answer:
m = 63 grams
Explanation:
ω = 10 cycles/s(2π radians/cycle) = 20π rad/s
ω = √(k/m)
m = k/ω² = 250/(20π)² = 0.06332... kg
t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec
In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)
d1 = 1*0.319 = 0.319 m
d2 = 0.5*0.451 = 0.225 m
in which d = Horizontal distance
ratio
= di : d2
= 0.319 : 0.225
= 3.19 : 2.25
Answer:
F = 683.8 N
Explanation:
The gravitational force of attraction between the Earth and the student is given by Newton's Law of Gravitation as follows:
where,
F = Force = ?
G = Universal gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
m₁ = mass of Earth = 5.98 x 10²⁴ kg
m₂ = mass of student = 70 kg
r = distance between Earth and student = 6.39 x 10⁶ m
Therefore,
<u>F = 683.8 N</u>
The resultant force will be towards Team A and its value will be:
450 - 415
= 35 Newtons
