The first time we read this, we think that we'll have to go look up a bunch of stuff about the gravitational constant, centripetal force, orbital mechanics, the Moon's period of revolution, and who knows whut owl !
But you know whut ? The only thing we need is Newton's 2nd law:
F = m A
Divide each side by m :
Acceleration = Force / mass
Acceleration = (2.03 x 10²⁰ Newtons) / (7.35 x 10²² kg)
Acceleration = (2.03 x 10²⁰ / 7.35 x 10²² kg) (m/s²)
<em>Acceleration = 0.00276 m/s²</em>
Answer:
average force = 385,140 N
Explanation:
from the question we are given the following
mass (m) = 1800 kg
distance of fall (d) = 3 m
driven distance (l) = 14.4 cm = 0.144 m
acceleration due to gravity (g) = 9.8 m/s^{2}
work done = average force x driven distance.....equation 1
and
work done = change in kinetic energy + change in potential energy
work done = (0.5 x m x (v^{2} - u^{2})) + (m x g x (-d-l))
- Initial velocity (u) and final velocity (v) are zero because the pile driver is it rest before it moves to hit the pile and after hitting the pile.
- The changes in length for the potential energy are negative because the pile moves downward
we now have work done = (m x g x (-d-l))...equation 2
now equating the two equations for work done we have
average force x driven distance = (m x g x (-d-l))
average force x 0.144 = 1800 x 9.8 x (-3-0.144)
average force = (1800 x 9.8 x (-3-0.144)) ÷ 0.144
average force = 385,140 N
Answer:
Secondary current is 8A
Explanation:
Data :
Primary supply = Vp = 240 V
Secondary supply = Vs = 60V
Primary current = Ip = 2A
Secondary current = Is = ?
Formula : Vp × Ip = Vs × Is
So for secondary current formula becomes
Is = (Vp/Vs)×Ip = (240/60) × 2
Is = 4 × 2 = 8A
<h2>
Relationship Between Frequency and Period</h2>
The frequency and the period are inversely proportional.
where T is the period
<h2>Solving the Question</h2>
We're given:
- <em>v</em> = 340 m/s
- <em>f</em> = 1000 Hz
Because the frequency and the period are reciprocals of each other, we can find the period of the sound by finding the reciprocal of the frequency:
<h2>Answer</h2>
