The final speed of the toy car at the end of the given time period is 3.58 m/s.
The given parameters;
- distance traveled by the car, s = 1.2 m
- time of motion of the car, t = 0.67 s
- initial velocity of the car, u = 0
The acceleration of the car is calculated as;
The final velocity of the toy car is calculated as;
Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.
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Answer:
v=32.49 m/s
Explanation:
Given that
Distance ,d= 66 m
Initial speed of the car ,u = 0 m/s
Coefficient of friction ,μ = 0.8
Lets take the total mass of the car = m
The acceleration of the car is given as
a = μ g ( g= 10 m/s² )
Now by putting the values in the above equation we get
a= 0.8 x 10 m/s²
a= 8 m/s²
We know that ,final speed is given as
v²= u ²+ 2 a d
Now putting the value
v²=0² + 2 x 8 x 66
v²= 1056
v=32.49 m/s
Answer:
As we need to use a nested loop in our function,hence push $ra
pop $ra
jal nested_function_label
nop is the correct option.
The correct answer is
C. reduce the friction between its moving parts
In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.
