Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
Answer:
The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid
B. The contact area of each tire with the track.
C. The co-efficent of static friction between the tires and the track.
D. The co-efficent of static friction between the tires and the track.
Explanation:
Answer:
Explanation:
Moment of inertia of larger disk I₁ = 1/2 MR²
Moment of inertia of smaller disk I₂ = 1/2 m r ²
Initial angular velocity
We shall apply law of conservation of angular momentum .
initial total momentum = final angular momentum
I₁ X ωi = ( I₁ + I₂ )ωf
1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf
ωf = ωi / ( 1 + m r²/MR² )
Answer:
The ball was in air for 3.896 s
Explanation:
given,
g = 9.8 m/s², acceleration due to gravity,
If the launch angle is 45°, the horizontal range will be maximum.
The horizontal and vertical launch velocities are equal, and each is equal to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The time to attain maximum height is one half of the time of flight.
v = u + at ∵ v = 0 (max. height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The time of flight is twice of the maximum height time
2 t₁ = 3.896 s
The horizontal distance traveled is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in air for 3.896 s
Answer:
We need 4 times more force to keep the car in circular motion if the velocity gets double.
Explanation:
Lets take the mass of the car = m
The radius of the arc = r
Given that speed of the car gets double ,v' = 2 v
Then the force on the car = F'
( radius of the arc is constant)
We know that
Therefore F' = 4 F
So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.
