λ=v/f
λ-wavelength
v-speed
f-frequency
we have the wavelength(6.2 x 10^-6meters) and we use the speed of light which is equal to 3*10^8m/s
6.2*10^-6m=3*10^8m/s/f
f=(3*10^8m/s)/(6.2*10^-6)≈0.48*10^14Hz
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
<u>FOR BULB 1:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
<u>FOR BULB 2:</u>
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
<u>A₁/A₂ = 0.44</u>
Answer:
You are exactly right. The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less
On ground weight of plane will be measured as its actual weight which will be given as
now when plane is in air its weight is measured as combined effect of earth gravitational force and buoyancy force both
so weight in air will be given as
here since net effect due to air is opposite to the direction of weight so in air the plane weight will be measure less than its weight on ground.
so answer will be
A)more than a plane in the air
Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force 
Mass of object = 2.00 kg
Initial position 
Final position 
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
Put the value into the formula
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.
