It's downward vector from the centre of apple in the direction to the centre of the Earth.
Answer:
The divergence on the sensor shows the magnitude of the charges
Explanation:
This will increase as length increases since it is said to be proportional to the length. note that test charge is always positive and charge on the grid is positive as indicated (1 nC)
Answer:
(a) the height in the open water column is: 18.15 (ft), (b) the gauge pressure acting on the bottom tank surface AB is: 8.73 (Psi) and (c) the absolute pressure of the air in the top of the tank with local atmospheric pressure at 14.7 Psi will be: 21.7 (Psi).
Explanation:
We need to apply the Pascal's Law, () where P is pressure, γ is the specific weight of the fluid and h is the height of the column of fluid, to find the pressure at differents points in the tank (see attached). First we need to assume a property of the water called specific weight as: 62.4 (lbf/ft^3) or 0.03612 (lbf/in^3). Then knowing that 1 feet as the same as 12 inches and appling Pascal's Law, we get: and the other hand, we have: and knowing that the pressure at point C and point D are the same, we can find the h as:, so getting h=217.75(in) or 18.15(ft) (a). Now similarly to find the gauge pressure acting on the bottom of the tank surface AB we can apply Pascal's Law as: (b). Finally we can find the absolute pressure of the air in the top of the tank, assuming a local atmospheric pressure as 14.7 (Psi) so: (c).
Answer:
(orbital speed of the satellite) V₀ = 3.818 km
Time (t) = 4.5 × 10⁴s
Explanation:
Given that:
The radius of the Earth is 6.37 × 10⁶ m; &
the acceleration of gravity at the satellite’s altitude is 0.532655 m/s
We can calculate the orbital speed of the satellite by using the formula:
Orbital Speed (V₀) = √(r × g)
radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m
= (2.1 × 10⁷ + 6.37 × 10⁶) m
= 27370000
= 2.737 × 10⁷m
Orbital Speed (V₀) = √(r × g)
Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )
= 3818.215
= 3.818 × 10³
= 3.818 Km
To find the time it takes to complete one orbit around the Earth; we use the formula:
Time (t) = 2 π ×
= 2 × 3.14 ×
= 45019.28
= 4.5 × 10 ⁴ s
2) all of their motion stops completely