Answer:
The specific heat is 3.47222 J/kg°C.
Explanation:
Given that,
Temperature = 13°C
Temperature = 37°C
Mass = 60 Kg
Energy = 5000 J
We need to calculate the specific heat
Using formula of energy
Put the value into the formula
Hence, The specific heat is 3.47222 J/kg°C.
Answer:
they meet from point o at distance 50.46 m and time taken is 11.6 seconds
Explanation:
given data
acceleration = 0.75 m/s²
speed B = 6 m/s
time B = 20 s
to find out
when and where the vehicles passed each other
solution
we consider here distance = x , when they meet after o point
and time = t for meet point z
we find first Bus B distance for 20 s ec
distance B = velocity × time
distance B = 6 × 20
distance B = 120 m
so
B take time to meet is calculate by distance formula
distance = velocity × time
120 - x = 6 × t
x = 120 - 6t .................1
and
distance of A when they meet by distance formula
distance = ut + 1/2 × at²
here u is initial speed = 0 and t is time
x = 0 + 1/2 × 0.75 × t²
x = 0.375 × t² .............2
so from equation 1 and 2
0.375 × t² = 120 - 6t
t = 11.6
so time is 11.6 second
and
distance from point o from equation 2
x = 0.375 (11.6)²
x = 50.46
so distance from point o is 50.46 m
Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as
Cp-Cv= R
Cp=7/2 R
Now by putting the values
a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that
Process is adiabatic
<span>The unknown substance is silver.
I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so
273 g / 26 mL = 10.5 g/mL
Looking up a list of elements sorted by density, I see the following:
10.07 Actinium
10.22 Molybdenum
10.5 Silver
11.35 Lead
And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
Explanation :
Absorption coefficient of a material determines how much sound is absorbed by the material.
To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.
Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.
Hence, option (A) and (D) are correct.
