Answer:
c neither
Explanation:
the cactus is just kinda chillin there
Answer:
Δt'/ T% = 90.3%
Explanation:
Simple harmonic movement is described by the expression
x = A cos (wt)
we find the time for the two points of motion
x = - 0.3 A
-0.3 A = A cos (w t₁)
w t₁ = cos -1 (-0.3)
remember that angles are in radians
w t₁ = 1.875 rad
x = 0.3 A
0.3 A = A cos w t₂
w t₂ = cos -1 (0.3)
w t₂ = 1,266 rad
Now let's calculate the time of a complete period
x= -A
w t₃ = cos⁻¹ (-1)
w t₃ = π rad
this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period
T = 2 t₃
T = 2π / w s
now we can calculate the fraction of time in the given time interval
Δt / T = (t₁ -t₂) / T
Δt / T = (1,875 - 1,266) / 2pi
Δt / T = 0.0969
This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is
Δt'/ T = 1 - 0.0969
Δt '/ T = 0.903
Δt'/ T% = 90.3%
The answer for this question is negative externality
Answer:
98N and 147N
Explanation:
We have the following information:
We can find the static fricton force as follow,
Where N is the normal force (mg)
Static friction force at 147N is greater than the force applied hence body does not move.
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
