Answer:
natural motion and violent motion
Explanation:
As we know that spring force is given as

here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have


so the elastic constant of the spring will be 200 N/m
Answer:
51 Ω.
Explanation:
We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:
Resistor 1 (R₁) = 40 Ω
Resistor 3 (R₃) = 70.8 Ω
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?
Since the two resistors are in parallel connection, their equivalent can be obtained as follow:
R₁ₙ₃ = R₁ × R₃ / R₁ + R₃
R₁ₙ₃ = 40 × 70.8 / 40 + 70.8
R₁ₙ₃ = 2832 / 110.8
R₁ₙ₃ = 25.6 Ω
Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω
Resistor 2 (R₂) = 25.4 Ω
Equivalent Resistance (Rₑq) =?
Rₑq = R₁ₙ₃ + R₂ (series connection)
Rₑq = 25.6 + 25.4
Rₑq = 51 Ω
Therefore, the equivalent resistance of the group is 51 Ω.
Answer:
Explanation:
Given that,
Bathysphere radius
r = 1.5m
Mass of bathysphere
M = 1.2 × 10⁴ kg
Constant speed of descending.
v = 1.2m/s
Resistive force
Fr = 1100N upward direction
Density of water
ρ = 1.03 × 10³kg/m³
The volume of the bathysphere can be calculated using
V = 4πr³ / 3
V = 4π × 1.5³ / 3
V = 14.14 m³
The Bouyant force can be calculated using
Fb = ρgV
Fb = 1.03 × 10³ × 9.81 × 14.14
Fb = 142,846.18 N
Buoyant force is acting upward
Weight of the bathysphere
W = mg
W = 1.2 × 10⁴ × 9.81
W = 117,720 N
Weight is acting downward
The net positive buoyant using resolving
Fb+ = Fb — W
Fb+ = 142,846.18 — 117,720
Fb+ = 25,126.18 N
The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force
W = Fb+ + Fr
W = 25,126.18 + 1100
W = 26,226.18
mg = 26,226.18
m = 26,226.18 / 9.81
m = 2673.4kg
Mass of submarine is 2673.4kg