Answer:
A.) 1.3 seconds
B.) 0.42 m
Explanation:
A.) You are given the angle of projection to be 40 degrees and initial velocity of 20m/s.
At vertical component
U = Usin 40 that is,
U = 20sin40
Using the first equation of motion under gravity
V = U - gt
Let V = 0
0 = UsinØ - gt
gt = UsinØ
t = UsinØ/g
Where U = 20 m/s
Ø = 40 degree
g = 9.8 m/s^2
Substitutes all the parameters into the formula
t = 20sin40/9.8
t = 1.3 seconds
Total time of flight T = 2t
T = 2 × 1.3 = 2.6 s
B.) To calculate the maximum height,
You will use the formula
V^2 = U^2 - 2gH
At maximum height, V = 0
2gH = Usin^2Ø
H = Usin^2Ø/ 2g
H = 20 sin^2(40) ÷ 2(9.8)
H = 8.2635/19.6
H = 0.42 m
progession
(a) = 4.19 x N, and its direction is towards .
(b) It must be placed inside a hollow shell.
Let, = 165 kg, = 465 kg, = 60 kg, and the distance between and is 0.340 m.
(a) Since is placed midway between and , then its distance to both masses is 0.170 m.
From the Newton's law of universal gravitation,
F =
Where all variables have their usual meaning.
Then,
a. = -
=
= 2.25 x N
= 6.44 x N
∴ = = 6.44 x - 2.25 x
= 4.19 x N
The net force exerted by the two masses on the 60 kg object is 4.19 x N.
(ii) // = // - //
= 6.44 x - 2.25 x
(iii) The direction of the net force is to the right i.e towards .
(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.
Salt water is more dense than freshwater.