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
Complete Question:
Andy and Charlie are riding on a merry-go-round. Andy rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Charlie, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, which of the following best describes Andy's angular speed?'
( ) twice Charlie’s
( ) impossible to determine.
( ) the same as Charlie’s
( ) half of Charlie’s
Answer:
The same as Charlie's
Explanation:
As a merry-go-round is a rigid body, all points in the rotating body must have the same angular velocity, i.e., thet must rotate the same angle in the same time.
Otherwise, the distance between any pair of points on a given radius could be different in different times, which is not possible in a rigid body,
Answer:
The light used has a wavelenght of 4.51×10^-7 m.
Explanation:
let:
n be the order fringe
Ф be the angle that the light makes
d is the slit spacing of the grating
λ be the wavelength of the light
then, by Bragg's law:
n×λ = d×sin(Ф)
λ = d×sin(Ф)/n
λ = (3.2×10^-4 cm)×sin(25.0°)/3
= 4.51×10^-5 cm
≈ 4.51×10^-7 m
Therefore, the light used has a wavelenght of 4.51×10^-7 m.
Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..
