Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin
1 answer:
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Answer:
15s
Explanation:
Given parameters:
Distance traveled = 375m
Speed = 25m/s
Unknown:
Time taken = ?
Solution:
To solve this problem, we make time the subject of the speed equation.
Speed =
Time =
Now insert the parameters and solve;
Time = = 15s
Answer:
The depth to which the hydrometer sinks is approximately 24.07 cm
Explanation:
The given parameters are;
The diameter of the hydrometer tube, d = 2.3 cm
The mass of the content of the tube, m = 80 g
The density of the liquid in which the tube floats, ρ = 800 kg/m³
By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid
When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer
Therefore;
The weight of the liquid displaced = The weight of the hydrometer, W = m·g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ W = 80 g × g
The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ
V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³
The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged,
= A × h
Where;
A = The cross-sectional area of the tube = π·d²/4
h = The depth to which the hydrometer sinks
h = /A
∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm
The depth to which the tube sinks, h ≈ 24.07 cm.
Answer:
Explanation:
plate separation = 2.3 x 10⁻³ m
capacity C₁ = ε A / d
= ε A / 2.3 x 10⁻³
C₂ = ε A / 1.15 x 10⁻³
=
a ) when charge remains constant
energy =
q is charge and C is capacity
energy stored initially E₁=
energy stored finally E₂ =
=
=
=
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C V²
energy is proportional to capacity as V is constant .
= 16.76 .
Answer:
2 /s north
Explanation:
Given that,
Velocity due North is 8 m/s and due south is 6 m/s
We need to find the magnitude and the direction of the resulting velocity.
Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,
It is positive. So, it is in North direction.