Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where =Mass of block B
Substitute the values
For block A
Where Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.
Hope this helps!
Answer:
d₁ = 0.29 in
d₂ = 0.505 in
Explanation:
Given:
T = 1500 lbf in
L = 10 in
x = 0.5 L = 5 in
First case: T = T₁ + T₂
T₂ = T - T₁ = 1500 - 750 = 750 lbf in
If the shafts are in series:
θ = θ₁ + θ₂
θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)
Second case: If d₁ ≠ d₂
θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)
t₁ = t₂
(eq. 2)
T₁ + T₂ = 1500 (eq. 3)
θ₁ first case = θ₁ second case
Replacing:
The same way to θ₂:
From equation 2, we have:
d₁ = 0.587 * d₂
From equation 3, we have:
d₂ = 0.505 in
d₁ = 0.29 in
Answer:
The length of the boomerang is 0.364 m
Explanation:
The moment of inertia is:
Where
md = 0.05 kg
mh = 0.12 kg
r = d/2 = 0.273/2 = 0.1365 m
The length of the boomerang is: