Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
The tectonic plates are made up of Earth's crust and the upper part of the mantle layer underneath. Together the crust and upper mantle are called the lithosphere. hope this helps :)
The rule to get the average speed is as follows:
average speed = average distance / average time
We are given that:
distance = 250 m
time = 110 sec
Substitute with the givens in the above equation to get the average speed as follows:
average speed = 250/110 = 25/11 meters/sec
Answer:
The speed is 15 km/h or 4.16 m/s.
Explanation:
A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which the speed of the boat in km / h? And in m / s?
Given that,
Distance, d = 90 km = 90000 m
Time, t = 6 hours = 21600 s
Speed = distance/time
or
So, the required speed is 15 km/h or 4.16 m/s.
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
