Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
<u>v₁ = 3.5 m/s</u>
Answer:
Period of motion is approximately 0.5447 seconds
Explanation:
We start by calculating the constant "k" of the spring which can be derived from the fact that an object of mass 12 g produced a stretch of 3.4 cm: (we write everything in SI units)
F = k * x
0.012 kg * 9.8 m/s^2 = k 0.034 m
k = 0.012 kg * 9.8 m/s^2 / (0.034 m)
k = 3.46 N/m
now we use the formula for the period (T) of a spring of constant k with a hanging mass 'm':
which in our case becomes:
Answer:
The work done will be 57.15 J
Explanation:
Given that,
Mass = 12 kg
Distance = 3 m
Force = 22 N
Angle = 30°
We need to calculate the work done
The work done is defined as,
Where, F = force
d = displacement
Put the value into the formula
Hence, The work done will be 57.15 J