Answer:
Explanation:
For sound level in decibel scale the relation is
dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .
Putting the given values
61 = 10 log I / 10⁻¹²
log I / 10⁻¹² = 6.1
I = 10⁻¹² x 10⁶°¹

intensity of sound of 5 persons


= 10log 5 x 10⁶°¹
= 10( 6.1 + log 5 )
= 67.98
sound level will be 67.98 dB .
Explanation:
In a heat exchange, the temperature change is inversely proportional to the specific heat capacity. Since substance A has a heat capacity that is much greater than that of substance B, the temperature change of substance A will be less than the temperature change of substance B. Therefore, the final temperature is closer to that of
than
.
Answer:
25 m/s
Explanation:
First we should define the variables
T=4
Dx = 100
ay=-9.8
ax=0
We can use formula 1 from the BIG 5
x=(v+v0)t/2
By plugging in our variables we can get 100=4(v+v0)/2
Which is 50=v+v0
v=v0 since horizontal acceleration always equals zero
so 2v0 = 50
v0 = 25
Answer:
KE = KE (incidental) - KE of emitted photons
or KE = h * f - Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 - 1.2 = 4.96 eV
Answer:
Average acceleration on first part of the chunk is given as

Average acceleration on second part of the chunk is given as

Explanation:
By momentum conservation along x direction we will have

so we have


also by energy conservation






by solving above equation we will have


Average acceleration on first part of the chunk is given as


Average acceleration on second part of the chunk is given as

