Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
<u>v₂ = 30 m/s</u>
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b)
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
<u>m₁ = 12600 kg</u>
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c)
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
<u>m₂ = 12600 kg</u>
The further an object is from the centre of a planet, the lower it's gravitational force. Uranus had 14 times as much mass as earth, but it's also a lot bigger than earth. So assuming an object is on the surface of Uranus, it would be really far away from the centre of Uranus, therefore the gravitational force is less.
Hope this helps!
Explanation:
The given data is as follows.
Magnetic field strength (B) = 0.65 T
Speed (v) = 2.3 m/s
Induced emf (E) = ?
Formula for emf induced at the ends of the rod of length L which is moving with a speed of v is as follows.
E = BvL
Putting the given values into the formula as follows.
= 
= 1.495 L .............. (1)
When magnetic field is changed to 
= 0.48 T
Now, we assume that the speed be 
to get the emf 
.
Then, 
= 3.11 m/s
Therefore, we can conclude that the speed v of the rod be adjusted to reestablish the emf induced between the ends of the rod at its initial value is 3.11 m/s.
I = pressure amplitude given = 0.2 W/m²
dB = decibel reading
decibel reading from the pressure amplitude is given as
dB = 10 log₁₀ (I/10⁻¹²)
inserting the values in the above equation
dB = 10 log₁₀ (0.2/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹.10¹²)
dB = 10 log₁₀ (2 x 10¹²⁻¹)
dB = 10 log₁₀ (2 x 10¹¹)
dB = 113.01 db
hence the decibel reading comes out to be 113.01 db
