When heat energy is transferred from direct contact between a warm and a cold object , it is known as heat transfer by conduction.
In conduction, the heat transfer takes place within an object or between two objects in contact until the temperature becomes uniform. this kind of heat transfer continues until the temperature at two ends between which the heat transfer take place , becomes equal. Heat transfer takes place from point at high temperature to point at lower temperature.
Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance
We need to calculate the focal length
Using formula of lens
put the value into the formula
The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens
Put the value in to the formula
We need to calculate the magnification
Using formula of magnification
The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.
I believe that you would weigh around 68 or 69 N, or 7 kilograms.
Answer: Your question is missing below is the question
Question : What is the no-friction needed speed (in m/s ) for these turns?
answer:
20.1 m/s
Explanation:
2.5 mile track
number of turns = 4
length of each turn = 0.25 mile
banked at 9 12'
<u>Determine the no-friction needed speed </u>
First step : calculate the value of R
2πR / 4 = πR / 2
note : πR / 2 = 0.25 mile
∴ R = ( 0.25 * 2 ) / π
= 0.159 mile ≈ 256 m
Finally no-friction needed speed
tan θ = v^2 / gR
∴ v^2 = gR * tan θ
v = √9.81 * 256 * tan(9.2°) = 20.1 m/s
