Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
Answer:there talking about the house
Answer:
The resistance is 
Explanation:
Given that,
Diameter of tube = 8.5 mm
Length = 8 cm
Resistivity = 2.5 m
We need to calculate the resistance
The resistance is equal to the product of the resistivity and length divided by the area of cross section .
In mathematical form,
Where, 
=resistivity
l = length
A = area of cross section
Put the value into the formula
Hence, The resistance is
