Answer:
m = 20,000 kg
Explanation:
Force, 
Acceleration of the shark, 
It is required to find the mass of the shark. Let m is the mass. Using second law of motion to find it as follows :
F = ma
Putting the value of F and a to find m
So, the shark's mass is 20,000 kg.
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h = 
h = 
h = 1.55 m
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
- M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
- Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)
=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
1.602177e-15 joules are in 10,000 electron volts
