Good morning.
We have:
Where
j is the unitary vector in the direction of the
y-axis.
We have that
We add the vector
-a to both sides:
Therefore, the magnitude of
b is
47 units.
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Answer:
Total internal reflection is going on. The refractive index of water is about 1.3, so sin 90/sin r=1/sinr=1.3. So the fish can only see objects outside the water within about 50 degrees of the vertical
