365gpa I don’t even know what it is but this is with 10 point and I need it
Answer:
The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons
Explanation:
We know that for two point charges of magnitude 
the magnitude of force between them is given by
where
is constant
is the separation between the charges
Initially when the charges are separated by 2.4 meters the force can be calculated as
Now when the separation is reduced to 0.7 meters the force is similarly calculated as
Applying value of the constant we get
Thus 
Answer:
W=-109.12 kJ/kg
Q=-76.34 kJ/kg
Explanation:
The needed work W we will calculate by using the work equation for polytropic process and the heat transfer Q we will calculate by using the energy balance equation.
Before the calculations we first need to determine the final temperature T2. We will do that by using the given initial temperature T1 = 10°C, the given initial p_1 = 120 kPa and final p_2 = 800 kPa pressure and the polytropic index n = 1.2. Before the calculation we need to express the temperature in K units.
T1 = 10°C + 273 K = 283 K
T2 = ((p_2/p_1)^(n-1)/n)* T1
T2 = 388 K
Now we can use the heat capacity C_v, = 0.3122 kJ /kg K and the temperatures T1 and T2 to determine the change in internal energy ΔU.
ΔU = C_v*(T2-T1)
ΔU = 32.78 kJ/kg
to determine the work we will also need the initial v1 and final v2 specific volume. The initial specific volume v1 we can determine from the ideal gas equation. For the calculation we will need the initial pressure p_1, temperature T1 and the specific gas constant R = 0.2081 kJ /kg K.
v1=R*T1/p_1
v1=0.4908 m^3/kg
For the final specific volume we need to replace the initial temperature and pressure with the final.
v2=R*T2/p_2
v2=0.1009 m^3/kg
The work W is then:
W=p_2*v2-p_1*v1/n-1
W=-109.12 kJ/kg
The heat transfer Q we can calculate form the energy balance equation. For the calculation we will need the calculated work W and the change in internal energy ΔU.
Q=W+ΔU
Q=-76.34 kJ/kg
Answer:
F=1.47 KN
Explanation:
Given that
Diameter of plate = 25 cm
Height of pool h = 3 m
We know that force can be given as
F= P x A
P=ρ x g x h
Now by putting the values
P=1000 x 10 x 3
P= 30 KPa
F= 30 x 0.049 KN
F=1.47 KN
So the force on the plate will be 1.47 KN.
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