Answer:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Answer:
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The atoms and molecules in it are in constant motion. The kinetic energy of such a body is the measure of its temperature. Potential energy is classified depending on the applicable restoring force. Gravitational potential energy – potential energy of an object which is associated with gravitational force
Answer: 288.8 m
Explanation:
We have the following data:
is the time it takes to the child to reach the bottom of the slope
is the initial velocity (the child started from rest)
is the angle of the slope
is the length of the slope
Now, the Force exerted on the sled along the ramp is:
(1)
Where 
is the mass of the sled and 
its acceleration
In addition, if we draw a free body diagram of this sled, the force along the ramp will be:
(2)
Where 
is the acceleration due gravity
Then:
(3)
Finding :
(4)
(5)
(6)
Now, we will use the following kinematic equations to find :
(7)
(8)
Where 
is the final velocity
Finding from (7):
(9)
(10)
Substituting (10) in (8):
(11)
Finding :
Answer:
a) T=1.35s
b) amplitude = 0.0923m
Explanation:
m=300 gr
k=6.5 N/m
first we need to get the angular frequency of the motion
so we have that
ω = √(k/m)
in this case motion is a simple harmonic so the period is defined by:
T= 2π / ω
T= 2π / √(k/m)
replacing the variables...
T= 2π / √(6.5/0.3)
T=1.35s (period of the block's motion)
and...
α max = | ω²r max |
2 = (2π/1.35)² * r max
r max= 0.0923m
