If I'm right you bring in newtons second law which is F=ma
you rearrange to a=F/m
so 18/5 (33-15 - unbalanced forces)
acceleration is 3.6m/s squared
Answer:
Explanation:
gauge pressure due to a liquid column of density d and height h is given by the following expression .
P = hdg
The pressure depends upon height of liquid column and not on the cross sectional area .
In first cylinder .
gauge pressure = .40 atm
hdg = .40 atm
cross sectional area of cylinder = π r²
The radius of second cylinder is twice of the first , cross sectional area will be 4 times .
The volume remains the same when the liquid is poured into second cylinder
volume = cross sectional area x height .
As cross sectional area of second cylinder is 4 times , height of liquid column in second cylinder = h / 4 .
gauge pressure in second cylinder = h / 4 x d x g = hdg / 4
.40 / 4 = .10 atm
gauge pressure in second cylinder = .10 atm.
Answer:
I think D. Since a vacuum iswhere gravity is concentrated. not sure
When plane is flying in tail wind condition we will have
Now during return journey we can say its would be head wind now
time = 2 + 1 = 2 hours
now add two equations
distance = 360 miles
now from above equation again
Answer:
1010 m
Explanation:
The following data were obtained from the question:
Height (h) of cliff = 500 m
Horizontal velocity (u) = 100 m/s
Horizontal distance (s) =?
Next we shall determine the time taken for the cannon ball to hit the ground. This can be obtained as follow:
Height (h) of cliff = 500 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
500 = ½ × 9.8 × t²
500 = 4.9 × t²
Divide both side by 4.9
t² = 500/4.9
Take the square root of both side
t = √(500/4.9)
t = 10.1 s
Finally, we shall determine the horizontal distance travelled by the cannon ball.
This can be obtained by using the following formula (s = ut) as illustrated below:
Horizontal velocity (u) = 100 m/s
Time (t) = 10.1 s
Horizontal distance (s) =?
s = ut
s = 100 × 10.1
s = 1010 m
Thus, the cannon ball was launched 1010 m away from the cliff.