Answer:
About 133 db.
Explanation:
Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.
I = 200 w / 10 m^2 = 20 w per square meter
I sub 0 = 10^-12 w per square meter
SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db
Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D
Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
Answer:
The amount of each gas that can dissolve in the ocean depends on the solubility and saturation of the gas in water. Solubility refers to the amount of a dissolved gas that the water can hold under a particular set of conditions, which are usually defined as 0o C and 1 atmosphere of pressure.
Explanation:
hope this helps
Answer:
117.72kW
Explanation:
Given data
Mass m= 50kg
height x = 2m
time taken = 2 minutes= 129 seconds
let us find the work done
WD= force * distance
WD= mgx
WD= 50*9.81*2
WD= 981 Joules
Let us find the power
Power= work * time
Power= 981*120
Power= 117720
Power= 117.72 kW
Hence the power spent is 117.72kW
Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%
25%
