Using the precise speed of light in a vacuum (
), and your given distance of
, we can convert and cancel units to find the answer. The distance in m, using
, is
. Next, for the speed of light, we convert from s to min, using
, so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is
6.405 min.
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
<h2>
Answer:</h2>
1000th multiple of the standard reference level for intensities.
<h2>
Explanation:</h2>
The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;
β = 10 log (I / I₀) --------------------(i)
Where;
I₀ = reference intensity
Given from the question;
β = sound level = 30dB
Substitute this value into equation (i) as follows;
30 = 10 log (I / I₀)
Divide both sides by 3;
3 = log (I / I₀)
Take antilog of both sides;
10^(3) = (I / I₀)
1000 = I / I₀
Solve for I;
I = 1000I₀
Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)
Answer:
Explanation:
Let the linear charge density of the charged wire is given as
here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have
here we have
so we have
