Answer:
Approximate escape speed = 45.3 km/s
Explanation:
Escape speed

Here we have
Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²
R = 1 AU = 1.496 × 10¹¹ m
M = 2.3 × 10³⁰ kg
Substituting

Approximate escape speed = 45.3 km/s
Answer:
Explanation:
Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.
Let T be tension in the hanging string
T cosθ = mg ( for balancing in vertical direction )
for balancing in horizontal direction
Tsinθ = F ( F is force of repulsion between two charges sphere)
Dividing the two equations
Tanθ = F / mg
tan17 = F / (7.1 x 10⁻³ x 9.8)
F = 21.27 x 10⁻³ N
if q be charge on each sphere , force of repulsion between the two
F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )
21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²
q² = .3973 x 10⁻¹²
q = .63 x 10⁻⁶ C
no of electrons required = q / charge on a single electron
= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .39375 x 10¹³
3.9375 x 10¹² .
You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz