Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
Explanation:
PEgrav = m *• g • h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.
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Potential Energy - The
Explanation:
A) Use Hooke's law to find the spring constant.
F = kx
40 N = k (0.4 m)
k = 100 N/m
B) Period of a spring-mass system is:
T = 2π √(m / k)
T = 2π √(2.6 kg / 100 N/m)
T = 1 s
Frequency is the inverse of period.
f = 1 / T
f = 1 Hz
Let the data is as following
mass of payload = "m"
mass of Moon = "M"
now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface
So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other
so it is given by
we know that
now from above formula
so above is the work done to move the mass from surface to given altitude
