Answer:
Tarzan will be moving at 7.4 m/s.
Explanation:
From the question given above, the following data were obtained:
Height (h) of cliff = 2.8 m
Initial velocity (u) = 0 m/s
Final velocity (v) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 2.8)
v² = 0 + 54.88
v² = 54.88
Take the square root of both side
v = √54.88
v = 7.4 m/s
Therefore, Tarzan will be moving at 7.4 m/s at the bottom.
Explanation:
The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through. The fault labeled "E" cuts through all three sedimentary rock layers (A, B,and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen and known of.
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
The net force is 12 N to the left.
