Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
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They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
Answer
given,
y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]
length of the rope = 1.33 m
mass of the rope = 3.31 g
comparing the given equation from the general wave equation
y(x,t)= A cos[k x+ω t]
A is amplitude
now on comparing
a) Amplitude = 2.20 mm
b) frequency =


f = 118.25 Hz
c) wavelength




d) speed


v = 105.84 m/s
e) direction of the motion will be in negative x-direction
f) tension


T = 27.87 N
g) Power transmitted by the wave


P = 0.438 W
Answer:
The force needed is the weight of the rock minus the buoyant force.
Explanation:
Answer:
-2040 m/s²
Explanation:
Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s
a = -2040 m/s²