Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
Answer:0.061
Explanation:
Given

Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by



integrating From
to 


![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by

Fraction of the total heat that is lost by the soup can be turned is given by

![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)




Answer:
density is
Mg/µL
Explanation:
given data
density of nuclear =
kg/m³
1 ml = 1 cm³
to find out
density of nuclear matter in Mg/µL
solution
we know here
1 Mg = 1000 kg
so
1 m³ is equal to
cm³
and here 1 cm³ is equal to 1 mL
so we can say 1 mL is equal to 10³ µL
so by these we can convert density
density =
kg/m³
density =
kg/m³ ×
Mg/µL
density =
Mg/µL
Answer:
She can make have 30 jars with raspberries in them with 50 left over.
Explanation:
1,700 divided by 55
30 equally
but 50 left over
This means that she can make have 30 jars with raspberries in them with 50 left over.
Answer:
a) S = 2.35 10³ J/m²2
,
b)and the tape recorder must be in the positive Z-axis direction.
the answer is 5
c) the direction of the positive x axis
Explanation:
a) The Poynting vector or intensity of an electromagnetic wave is
S = 1 /μ₀ E x B
if we use that the fields are in phase
B = E / c
we substitute
S = E² /μ₀ c
let's calculate
s = 941 2 / (4π 10⁻⁷ 3 10⁸)
S = 2.35 10³ J/m²2
b) the two fields are perpendicular to each other and in the direction of propagation of the radiation
In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.
the answer is 5
C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis