Answer: car B has travelled 4times as far as Car A
d=vi*t+1/2at^2
No initial velocity so equation becomes;
d=1/2at^2 and the acceleration is the same between both only time is different;
Car A d=1/2a(1)^2
Car B d=1/2a(2)^2
Car A d= 1^2=1
Car B d= 2^2=4
Car B d=4*Car A
So car B has travelled 4 times as far as car A
Work = force x distance
= 100N (force) x 0.5m (distance)
= 50J
Velocity = displacement/time
I hope i helped you <33
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.
From Newton's second law we understand that
Gravity at this case)
Where,
m = mass
a= acceleration
Also we know that
Part A) The buoyant force acting on the balloon is given as
As mass is equal to the density and Volume and acceleration equal to Gravity constant
PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then
PART C) The additional mass that can the balloon support in equilibrium is given as
