Answer:
I think it is Velocity idk but does it hurt to try
Answer:
Coefficient of friction is
Ū = 0.31
Explanation:
T2 = T1* e^(ūơ)
Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,
ū = coefficient of friction
Ơ = 2pai * N
Where N = number of turns = 2
Ơ = 4pai
7500 = 150e^(ū*4pai)
50 = e^(ū *4pai)
lin 50 = 4pai * ū
Ū = 3.91/4pai
Ū = 0.31
The highest elevation reached by the ball in its trajectory is 16.4 m.
To find the answer, we need to know about the maximum height reached in a projectile.
What's the mathematical expression of the maximum height reached in a projectile motion?
- The maximum height= U²× sin²(θ)/g
- U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity
What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?
- Here, U = 30.0 m/s and θ= 25°
- Maximum height= 30²× sin²(25)/9.8
= 16.4m
Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.
Learn more about the projectile motion here:
brainly.com/question/24216590
#SPJ4
Answer:
Largest
[q1=+1nC, q2=-1nC, q3=-1nC] & [q1=-1nC, q2=+1nC, q3=+1nC]
[q1=+1nC, q2=-1nC, q3=+1nC]
[q1=+1nC, q2=+1nC, q3=-1nC]
Smallest
[q1=+1nC, q2=+1nC, q3=+1nC] & [q1=-1nC, q2=-1nC, q3=-1nC]
Explanation:
Answer:
maximum amplitude = 0.08 m
Explanation:
Given that
Time period T= 0.58 s
acceleration of gravity g= 9.8 m/s²
We know that time period of simple harmonic motion given as
T = 2π/ω
0.58 = 2π/ω
ω = 10.83rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
9.8 = 10.83² A
A = 0.08m
maximum amplitude = 0.08 m