Answer:
=0.855V
Explanation:
The induced voltage can be calculated using below expression
E =B x dA/dt
Where dA/dt = area
B= magnetic field = 6.90×10-5 T.
We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit
speed = 885 km/h
speed = 885 x 10^3 m/hr
speed = 885 x 10^3/60 x60 m/s
speed = 245.8 m/s
If The aircraft wing sweep out" an area
at t= 50.4seconds then we have;
dA/dt = 50.4 x 245.8
= 123388.32m^2/s
Then from the expression above
E =B x dA/dt substitute the values of each parameters, we have
E = 6.90 x 10^-5 x 12388.32 V
E =0.855V
Hence, the average induced voltage between the tips of the wings is =0.855V
Aromatherapy based treatment can offer impermanent help when nasal and sinus tissue drainage blockage and pressure for Mike.
<u>Explanation:</u>
Nasal and sinus tissues provide stuffy nose and pain or pressure in several areas around the face or teeth. There's normally a nasal release that might be yellow, green, or clear. You may likewise have exhaustion, the issue with a feeling of smell or taste, hack, sore throat, terrible breath, migraine, torment when you twist forward, and fever.
For this issue, Mike can utilize Aromatherapy based treatment that can offer some quick help with discomfort on the blocked nose and migraine. During this stage, mike ought to breathe in some restorative or common smell to control his sufferings. Fragrance, for example, Peppermint, Eucalyptus, Tea tree, Oregano, Clary wise, Lavender, Rosemary, and so forth are a great idea to breathe in.
Answer:
241 kPa
Explanation:
The ideal gas law states that:

where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can rewrite the equation as

For a fixed amount of gas, n is constant, so we can write

Therefore, for a gas which undergoes a transformation we have

where the labels 1 and 2 refer to the initial and final conditions of the gas.
For the sample of gas in this problem we have

So we can solve the formula for
, the final pressure:

<span>d. The parallaxes beyond a few thousand light years are
too small to be measured with common instruments.
I'm not sure that parallax can even be used out to a few
thousand light years.
The NEAREST star to Earth has the BIGGEST parallax.
The star is Alpha Centauri. It's only 4 light years away
from us, and its parallax is 0.000206 of a degree !
I have no idea how astronomers can measure angles
so small ... and that's the BIGGEST parallax angle of
ANY star.</span>
Answer:

Explanation:
The radiated power can be given in terms of the wavelength as follows:

where,
Radiated Power = 1.2 x 10⁻¹⁷ W
n = no. of photons = ?
h = plank's constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 500 nm = 5 x 10⁻⁷ m
t = time
Therefore,
