Answer:
x = 45 cm
Explanation:
Given that,
The length of a rod, L = 50 cm
Mass, m₁ = 0.2 kg
It is at 40cm from the left end of the rod.
We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.
The centre of mass of the rod is at 25 cm.
Taking moments of both masses such that,

The distance from the left end is 40+5 = 45 cm.
Hence, at a distance of 45 cm from the left end it will balance the rod.
Answer:
I believe the answer is

Explanation:
I may be incorrect because I used to do this a long time ago but I believe I am correct
HOPE THIS HELPS!
(:
Answer:
Δ T = 2.28°C
Explanation:
given,
mass of marble = 100 Kg
height of fall = 200 m
acceleration due to gravity = 9.8 m/s²
C_marble = 860 J/(kg °C)
using conservation of energy
Potential energy = heat energy
Δ T = 2.28°C
Answer:
The sound level of the 26 geese is 
Explanation:
From the question we are told that
The sound level is 
The number of geese is 
Generally the intensity level of sound is mathematically represented as
The intensity of sound level in dB for one goose is mathematically represented as
![Z_1 = 10 log [\frac{I}{I_O} ]](https://tex.z-dn.net/?f=Z_1%20%3D%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_O%7D%20%5D)
Where I_o is the threshold level of intensity with value 
is the intensity for one goose in 
For 26 geese the intensity would be

Then the intensity of 26 geese in dB is
![Z_{26} = 10 log[\frac{26 I }{I_o} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%5B%5Cfrac%7B26%20I%20%7D%7BI_o%7D%20%5D)
![Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%2A%20%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
![Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%20%5C%20%5C%20%29%20%2A%20%20%20%28%5C%20%5C%20%2010%20log%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
From the law of logarithm we have that
![Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%2026%20%2B%20%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_0%7D%20%5D)


Answer:
=118.8 K= 154.2°C
Explanation:
COP_max of carnot heat pump= 
where T_H and T_C are temperatures of hot and cold reservoirs
Also COP=
in the question 
⇒
heat is added directly to be as efficient as via heat pump

and T_H= 24° C= 297 K

on calculating the above equation we get
=118.8 K
the outdoor temperature for efficient addition of heat to interior of home
=118.8 K= 154.2°C