Answer:
7.894 Hours.
Explanation:
Based on information number hours that this battery will last with give load has mathematical relation of.
with load 60A t = 1h, 30A t = 2h so on and forth.
two head lights draw total current of 2x3.8A = 7.6A.
putting this in above relation gives.
.
That is how long will it be before battery is dead.
Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
Answer:
0.24 kgm²
Explanation:
= length of the bat = 81.3 cm = 0.813 m
= mass of the bat = 0.96 kg
= distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m
= Period of oscillation = 1.35 sec
= moment of inertia of the bat
Period of oscillation is given as
= 0.24 kgm²
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m