Answer:
a = 0.63 m/s²
Explanation:
given,
mass of submarine = 1460-kg
upward buoyant force = 16670 N
downward resistive force = 1150 N
submarine acceleration = ?
assuming g = 10 m/s²
now,
B - (R + mg) = ma
16670 - 1150 - 1460 × 10 = 1460 × a
1460× a = 920
a = 0.63 m/s²
hence, the acceleration of submarine is equal to a = 0.63 m/s²
In component form, the displacement vectors become
• 350 m [S] ==> (0, -350) m
• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m
(which I interpret to mean 20° north of east]
• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m
Then the student's total displacement is the sum of these:
(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m
≈ (280.371, 328.452) m
which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].
Answer:
W = 1222.4 J = 1.22 KJ
Explanation:
The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:
W = F d Cosθ
where,
W = Work Done = ?
F = Force Applied = 64 N
d = Distance Covered by Box = 19.1 m
θ = Angle between force and displacement = 0°
Therefore,
W = (64 N)(19.1 m)Cos 0°
<u>W = 1222.4 J = 1.22 KJ</u>
E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N