Answer:
T=26.03 N
Explanation:
Given that
Distance between poles = 12 m
Mass of block m= 4 kg
Sag distance = 5 m
Lets take tension in the clothesline is T.
The component of tension in vertical direction will be T cosθ.
By force balancing
2 T cosθ = 40
here
θ=39.80°
2 T cos39.8 = 40
T=26.03 N
Answer:
The total distance is 381.5 [m]
Explanation:
In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
a = acceleration = 2 [m/s²]
t = time = 7 [s]
Vf = 0 + (2*7)
Vf = 14 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
14² = 0 + (2*7*x)
x = 196/(14)
x = 14 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.
We must calculate the final velocity.
Vf = final velocity [m/s]
Vi = initial velocity = 14 [m/s]
a = desacceleration = 4 [m/s²]
t = time = 8 [s]
Vf = 24 + (4*8)
Vf = 56 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
56² = 14² + (2*4*x)
x = 2940/(8)
x = 367.5 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].
By equation of motion we have v = u + at
Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration
Here v = 141 m/s, u = 17.7 m/s and t = 6 s
On substitution we will get
141 = 17.7+ 6a
So, a = (141-17.7)/6 = 20. 55 m/
Aceeleration = 20. 55 m/ along north direction.