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:
2.4 m/s". 1
Explanation:
A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1
Ok so we know:
The time (t) is 18seconds
The acceleration (a) is 2.2m/s2
The displacement (r) is 660
Using the equation

With 'u' being the initial velocity we want, we get:

So:

So:

So the original/initial velocity was 16.8666 or 16.87 m/s
Hope this helped