an eagle has a mass of 8 kg and is flying 85 m above he ground with a speed of 20 m/s. what is the total mechanical energy of th
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Answer:
Δt = 5.29 x 10⁻⁴ s = 0.529 ms
Explanation:
The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.
where,
Δt = required time interval = ?
Δs = distance between ears = 18 cm = 0.18 m
v = speed of sound = 340 m/s
Therefore,
<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>
Answer:
A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.
B) The Diameter of this spherical space station should be 30.4m
Explanation:
The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:
If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.
The volume of a sphere is:
So the diameter is:
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
(t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
(t) = V₀√(C/L)
we substitute
(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
(t) = 25 × 0.00002
(t) = 0.0005 A
(t) = 0.5 mA
Therefore, the inductor current is 0.5 mA