Answer:
v_avg = 37 km/h
Explanation:
To find the average velocity in the complete trajectory you use the following formula:
( 1 )
v1: velocity in the first part of the trajectory = 70 km/h
v2: velocity in the second part of the trajectory = ?
You calculate v2 by using the following equation for a motion with constant velocity:
you replace the values of v1 and v2 in (1) and you obtain:
hence, the average velocity is 37 km/h
Answer:
Maximum permitted surface crack length is 1.29 mm
Explanation:
As per the question:
Tensile stress,
Surface energy of magnesium oxide, SE =
Modulus of elasticity of the material, E = 225 GPa =
Now,
To calculate the maximum allowable surface crack length:
Explanation:
It is given that,
The acceleration of a particle, (negative as the particle is decelerating)
Initial distance, x₁ = 20 m
Initial time, t₁ = 4 s
New distance x₂ = 4 m
Velocity, v = 10 m/s
(A) Calculating initial distance using second equation of motion as :
u = 21 m/s
When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :
t = 2.62 seconds
So, the velocity of the particle is zero at t = 2.62 seconds.
(B) Velocity at t = 11 s
v = 13 m/s
Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.
d = 132.48 m
So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.