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:
96 Joules
Explanation:
The formula for work is Fnet times displacement (F x d = w) which, in this case, 48N is the Fnet and 2m as the displacement. Then all we need to do is multiply these two and we get 96 Joules.
Answer:
tension in rope = 25.0 N
Explanation:
- Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
- Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.
∑F = F - W = 0
so
F = W
so tension in rope = F = T = 25 N
ANSWER: d) 8
EXPLANATION: Two sets of two shared electrons (4 electrons total shared) = one set of a double covalent bond.
Therefore, 8 electrons total shared = two sets of double covalent bonds
