The average speed is determined by the following formula:
average speed = [sum of (speed * time for which that speed was traveled)] / total time
average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division
average speed = 50.65 km/hr
The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60
Distance = 116.5 kilometers
Explanation:
speed of light= c
wave length= L
frequency= f
c=Lf → L= c/f → L= 3 × 10⁸/ 27 × 10⁹ → L = 1/90 ≈ 0.011 m
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Answer:
(a) Bus will traveled further a distance of 40 m
(b) It will take 7.5 sec to stop the bus
Explanation:
We have given initial velocity of the bus u = 24 m/sec
And final velocity v = 16 m/sec
Distance traveled in this process s = 50 m
From third equation of motion we know that 
(a) Now as the bus finally stops so final velocity v = 0 m/sec
So
s= 90 m
So further distance traveled by bus = 90-50 =40 m
(b) Now as the bus finally stops so final velocity v= 0 m/sec
Initial velocity u = 24 m/sec
Acceleration
So time 
