Answer:
(a) The height of the cliff is 9.92 m
(b) Time taken to reach the ground is 0.81 s
Explanation:
Given;
time of motion of the rock, t = 2.48 s
initial velocity of the rock, u = 8.19 m/s
(a) This is a two-way problem;
maximum upward height reached by the rock (h) + Total distance traveled downwards from the top (H).
Height of cliff = H - h
Maximum height reached by the rock;
Time to reach the maximum height;
The time taken for rock to falll to the gound;
t₂ = 2.48 - 0.83 s
t₂ = 1.65 s
The distance traveled downwards is calculated as
H = vt + ¹/₂gt²
where;
v is the final velocity at maximum height = initial velocity when the rock starts moving downwards = 0
H = ¹/₂gt²
H = 0.5 x 9.8 x (1.65)²
H = 13.34 m
The height of the cliff = 13.34 m - 3.422m = 9.92 m
(b) time taken to reach the ground if thrown with the same speed;
h = ut + ¹/₂gt²
9.92 = 8.19t + (0.5 x 9.8)t²
9.92 = 8.19t + 4.9t²
4.9t² + 8.19t - 9.92 = 0
a = 4.9, b = 8.19, c = -9.92
use the quadratic formula;
Answer:
4 m/
Explanation:
From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.
The Net force = Mass * Acceleration.
Since Net Force = 20 Newton, Mass = 5kg, therefore;
20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;
Acceleration = which gives 4.
Note: RHS means Right Hand Side.
LHS means Left Hand Side.
Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then
Solving for we get
Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is
The exhaust rate dm/dt is
Therefore, the velocity at which the exhaust gases exit the engines is
Answer
given,
V = 2 L
the left is an ideal gas at P = 100 k Pa and T = 500 K
mass is constant
Pressure is same because it's not changing due to process
m = 1.39 x 10⁻³ Kg