Let h = distance (m) to the water surface.
Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)
h = (1/2)*(9.8)*(3^2) = 44.1 m
The correct response that will be used to describe this particular element would be the third option, since all of the other options are incorrect and apply to different elements in their groups. The element is a metal and will react with a non metal.
Answer:
the length of the simple pendulum is 0.25 m.
Explanation:
Given;
mass of the air-track glider, m = 0.25 kg
spring constant, k = 9.75 N/m
let the length of the simple pendulum = L
let the frequency of the air-track glider which is equal to frequency of simple pendulum = F
The oscillation frequency of air-track glider is calculated as;
The frequency of the simple pendulum is given as;
Thus, the length of the simple pendulum is 0.25 m.
Answer:
Explanation:
Since work done is in the form of potential energy, we will use the formula of potential energy here.
We know that,
<h3>P.E. = mgh </h3>
Where,
m = mass = 20 kg
g = acceleration due to gravity = 10 m/s²
h = vertical height = 20 m
So,
<h3>Work done = mgh</h3>
Work done = (20)(10)(20)
Work done = 4000 joules
Work done = 4 kJ
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Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
