Answer:
16. c
17. d
Explanation:
16. Newton’s third law states that a force will always have an opposite but equal force as a reaction, so every force comes as a pair of action-reaction forces. For example, if you push on a book, the book also pushes on you.
17. If 1cm represents 15N, then 5cm must represent 15N*5=75N.
Answer:
<em>Heat of the reservoir is 461.38 K or 188.1 °C</em>
<em>The heating load is 18.705 kW</em>
Explanation:
COP = 8.7
working temperature
= 248 °C = 248 + 273.3 = 521.3 K
work power W = 2.15 kW
reservoir temperature
= ?
heating load Q = ?
We know that
COP = Q/W
Q = COP x W = 8.7 x 2.15 = <em>18.705 kW</em>
Also,
COP =
= 
8.7 = 
4535.31 - 8.7
= 521.3
4535.31 - 521.3 = 8.7
4014.01 = 8.7
= 4014.01/8.7 = <em>461.38 K</em>
or 461.38 -273.3 = <em>188.1 °C</em>
Answer:
Maximum speed, v = 36 m/s
Explanation:
Given that,
The radius of the curved road, r = 120 m
Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :

g is the acceleration due to gravity


v = 36.13 m/s
or
v = 36 m/s
So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.
Answer:
Explained below
Explanation:
For the geometric meaning, I've attached a parallelogram showing two vectors a and B and also the angle between them as θ.
We can see that the two vectors are adjacent to each other.
So the magnitude of the product of the two vectors is simply the area of the parallelogram which is base x height. Thus;
Magnitude = |a| sin θ × |b|
This can be rewritten as;
|a|•|b| sin θ
Answer:
Mass of ice per second melt is 2.74×10^-5Kg/s
Explanation:
Temperature of one end of the copper rod is 100°C boiling point of water and the other end of the rod is 0°C
Temperature difference in the copper rod = 100 - 0 = 100°C
Cross sectional area = 3.6×10^-4m^2
Length of rod , L = 1.7m
Amount of heat transfer from the boiling water to the ice water mix through the copper rod is given by:
Q = KA◇T/ L
Q = (390×(3.6×10^-4)×100°C)/1.7
Q = 14.04/1.7
Q = 8.26J/s
From the equation
Q = mLf
m = Q/ Lf
Where Lf = Latent heat of fusion for water= 3.34×10^5J/Kg
m = 8.26/(3.34×10^5)
m = 2.74×10^-5Kg/s