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:
<em>The object could fall from six times the original height and still be safe</em>
Explanation:
<u>Free Falling</u>
When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.
If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

Where v is the speed just before hitting the ground
If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

Solving for h

If the drop had occurred in the Moon, then

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g


This means the object could fall from six times the original height and still be safe
Aluminium, or heavier version copper.
Answer:
Explanation:
Given
Two block are connected by rope 
rope is attached to block 2
suppose
is a force applied to Rope
Applied force
=Tension in Rope 2

where a=acceleration of system
Tension in rope
is denoted by 

divide 1 and 2 we get

also 



The gravitational force between two masses m₁ and m₂ is

where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm