I included things
Hope this helps!
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 distance = 400+700+1200= 2300km
Explanation:
the resultant of d 1st right angle triangle + 1200
= 806.2 + 1200 = 2006.2km
Answer:
Part a)
Velocity = 6.9 m/s
Part b)
Position = (3.6 m, 5.175 m)
Explanation:
Initial position of the particle is ORIGIN
also it initial speed is along +X direction given as

now the acceleration is given as

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero
so we will have



Part a)
the velocity of the particle at this moment in Y direction is given as



Part b)
X coordinate of the particle at this time



Y coordinate of the particle at this time



so position is given as (3.6 m, 5.175 m)
I think (50 joules ) I’m not sure