Answer:
6.32m/s
Explanation:
note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²
okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!
when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)
now enough of the talking let solve....
so the ball was dropped .ie from rest to the ground through a distance of 2m,
the formula for calculating the distance if a body moving in a straight line is given by:
S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.
from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)
solving ....
we get t to be 0.632s
to find the speed we substitute t in the equation below:
V=u+at ,but since u=0
V=at =10•0.632=6.32m/s
therefore the speed the body uses to strike the ground is 6.32m/s
Answer:
<h2>0.2 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
![a = \frac{f}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bf%7D%7Bm%7D%20%20%5C%5C%20)
f is the force
m is the mass
From the question we have
![a = \frac{100}{500} = \frac{1}{5} = 0.2 \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B100%7D%7B500%7D%20%20%3D%20%20%5Cfrac%7B1%7D%7B5%7D%20%20%3D%200.2%20%5C%5C%20)
We have the final answer as
<h3>0.2 m/s²</h3>
Hope this helps you
The frequency of a wave is the reciprocal of its period.
A period of 0.008 sec means a frequency of
1 / 0.008 sec = 125 per sec . (125 Hz)
When Jane is sliding down a slide, she is demonstrating translational motion.
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz