Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :
So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
The acceleration of the particle at time t is:
The velocity of the particle at time t is given by the integral of the acceleration a(t):
and the position of the particle at time t is given by the integral of the velocity v(t):
Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
Answer:
Tendons connect muscle to bone. These tough, yet flexible, bands of fibrous tissue attach the skeletal muscles to the bones they move. Essentially, tendons enable you to move; think of them as intermediaries between muscles and bones.
Hope this helps! (:
Answer:
Vf= 7.29 m/s
Explanation:
Two force act on the object:
1) Gravity
2) Air resistance
Upward motion:
Initial velocity = Vi= 10 m/s
Final velocity = Vf= 0 m/s
Gravity acting downward = g = -9.8 m/s²
Air resistance acting downward = a₁ = - 3 m/s²
Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²
( Acceleration is consider negative if it is in opposite direction of velocity )
Now
2as = Vf² - Vi²
⇒ 2 * (-12.8) *s = 0 - 10²
⇒-25.6 *s = -100
⇒ s = 100/ 25.6
⇒ s = 3.9 m
Downward motion:
Vi= 0 m/s
s = 3.9 m
Gravity acting downward = g = 9.8 m/s²
Air resistance acting upward = a₁ = - 3 m/s²
Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²
Now
2as = Vf² - Vi²
⇒ 2 * 6.8 * 3.9 = Vf² - 0
⇒ Vf² = 53. 125
⇒ Vf= 7.29 m/s
