Radiation: transferred energy
Convection: Transfer of heat
Conduction: transfer of electric charge
Answer: 13 m/s
Explanation: The two vectors form a 5-12-13 right triangle. the magnitude of their resultant is the hypotenuse, which is 13 m/s.
Answer:
The polytropic exponent (n) is 1.2
The amount of energy transfer by work is 552.45 Btu
Explanation:
P1V1^n = P2V2^n
P1 = 75 lbf/in^2 = 75 lbf/in^2 × 143.3 in^2/ft^2 = 10,747.5 lbf/ft^2
V1 = 4 ft^3/lb × 10 lb = 40 ft^3
P2 = 20 lbf/in^2 = 20 × 143.3 = 2866 lbf/ft^2
V2 = 12 ft^3/lb × 10 lb = 120 ft^3
(V2/V1)^n = P1/P2
(120/40)^n = 10747.5/2866
3^n = 3.75
ln 3^n = ln 3.75
n ln 3 = ln 3.75
n = ln 3.75/ln 3 = 1.2
W = (P2V2 - P1V1)/1 - n = (2866×120 - 10747.5×40)/1-1.2 = (343,920 - 429,900)/-0.2 = -85,980/-0.2 = 429,900 lbf.ft = 429,900/778.17 Btu = 552.45 Btu
The mass of an object that accelerates 1.5m/s² when a force of 7.0 newtons is applied to it is 4.67kg.
<h3>How to calculate mass?</h3>
The mass of an object can be calculated by dividing the force applied to the object by its acceleration. That is;
Mass = Force ÷ acceleration
According to this question, an object accelerates 1.5m/s² when a force of 7.0 newtons is applied to it. The mass of the object can be calculated as follows:
Mass = 7N ÷ 1.5m/s²
Mass = 4.67kg
Therefore, the mass of an object that accelerates 1.5m/s² when a force of 7.0 newtons is applied to it is 4.67kg.
Learn more about mass at: brainly.com/question/19694949
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Answer: 13.15m/s, N 20.052°W
Explanation: The diagramatic representation of this question has been attached to this answer, please kindly find below because references will be made to it.
Due to the fact that the boat is moving northwest with a speed of 14m/s and the speed of the water flows at 4.8m/s due north, they can be represented using triangular vectors.
From the attachment below.
H = 14m, O= 4.8m and A=?
Using phythagoras theorem, we have that
H² =O² + A²
14² = 4.8² + A²
A² = 14² - 4.8²
A² = 144 - 23.04
A² = 120.96
A= √120.96
A= 13.15m
The direction of the vector is gotten by using the fact that
tan θ = O/A
tan θ = 4.8/14
tan θ = 0.3650
θ = tan^-1 (0.3650)
θ= 20.052°
From the attachment, the vector is lying north east west, hence direction is N 20.052° W