Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
Convection means that hotter and less dense fluids have a tendency to rise while colder and more dense fluids sink.
The answer would be (A)
Hot water is denser than cold water and so hot water will be above the cold water.
:D
Answer:
(a) 
(b) 142
(c) 
(d) 96.8 mph
(e) 0.426 s
(f) 0.061 rad
Explanation:
Velocity is a time-derivative of position.
(a) 
(b) Since is independent of 
, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.
(c) 
(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to 
.
.
In this time, the vertical velocity, 
is
The speed of the ball at thus point is 
ft/s
To convert this to mph, we multiply the factor 3600/5280
(e) The time has been determined from (d) above.
(f) This angle is given by
(Note here we are considering the acute angle so we ignore the negative sign)
In radians, this is
Answer:
V₂ = 1.5 m/s
Explanation:
given,
speed of the first piece = 6 m/s
speed of the third piece = 3 m/s
speed of the second fragment = ?
mass ratios = 1 : 4 : 2
fragment break fly off = 120°
α = β = γ = 120°
sin α = sin β = sin γ = 0.866
using lammi's theorem
A,B and C is momentum of the fragments
4 x V₂ = 2 x 3
V₂ = 1.5 m/s
