Atoms seldom exist as independent particles in nature because...
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Answer:
(i) Please find attached the required velocity time graphs plotted with MS Excel
(ii) The velocity of vehicle A at the 18th second = 20 m/s
The velocity of vehicle B at the 18th second = 0 m/s
(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters
Explanation:
The given parameters of the motion of vehicles A and B are;
The acceleration of vehicles A and B = Uniform acceleration starting from rest
The maximum velocity attained by vehicle A = 30 m/s
The time it takes vehicle A to attain maximum velocity = 10 s
The maximum velocity attained by vehicle B = 30 m/s
The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s
The time duration vehicle A maintains its maximum velocity = 6 s
The time duration vehicle B maintains its maximum velocity = 4 s
(i) From the question, we get the following table;
From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here
(ii) The velocity of vehicle A at the start = 0 m/s
After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s
The maximum velocity is maintained for 6 seconds which gives;
At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s
The time it takes vehicle A to decelerate to rest = 6 s
The deceleration of vehicle A, = (30 m/s - 0 m/s)/(6 s) = 5 m/s²
Therefore, we get;
v = u - ·t
At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s
u = 30 m/s
∴ v₁₈ = 30 - 5 × 2 = 20
The velocity of vehicle A at the 18th second, = 20 m/s
For vehicle B, we have;
At the 14th second, the velocity of vehicle B = 40 m/s
Vehicle B decelerates to rest in, t = 4 s
The deceleration of vehicle B, = (40 m/s - 0 m/s)/(4 s) = 10 m/s²
For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s
∴ = 40 m/s - 10 m/s² × 4 s = 0 m/s
The velocity of the vehicle B at 18th second, = 0 m/s
(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;
The area triangle A₁ = (1/2) × 10 × 30 = 150
Area of rectangle, A₂ = 6 × 30 = 180
Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50
The distance covered in the 18th second by vehicle = A₁ + A₂ + A₃
∴ = 150 + 180 + 50 = 380
The distance covered in the 18th second by vehicle = 380 m
The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;
Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440
The distance covered by the trapezoid, = 440 m
The distance of the two vehicles apart at the 18t second, =
-
∴ = 440 m - 380 m = 60 m
The distance of the two vehicles from one another at the 18th second, = 60 m.
Answer:
The time taken is
Explanation:
From the question we are told that
The length of steel the wire is
The length of the copper wire is
The diameter of the wire is
The tension is
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
Where is the time taken to transverse the steel wire which is mathematically represented as
here is the density of steel with a value
So
And
is the time taken to transverse the copper wire which is mathematically represented as
here is the density of steel with a value
So
So