Answer:
57 kg
Explanation:
Mass of seesaw = 20 kg
Length of seesaw = 4 m
Mass of child on the longer end = 30 kg
The weight of the seesaw acts at the center i.e. 2m
The algebraic sum of moments of all forces about any point is zero, hence, using the fulcrum as the reference point:
[x * 9.8* 1.5] - [20 * 9.8* (2.5 - 2)] - [30 * 9.8 * 2.5] = 0
=> 14.7x = (20*9.8*0.5) + 735
14.7x = 98 + 735
14.7x = 833
=> x = 833/14.7
x = 57 kg
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
Area because CM^2 is the unit used for Area
Answer:
W = 2.3 10²
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = π r³
V = π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10²
therefore the weight of the drop is much greater than the electric force