In 1842, Julius Robert Mayer discovered The law of conservation of Energy. It its most compact form, it it now called The first law of Thermodynamics
Energy can neither be created nor destroyed, it can only be changed to another form of energy.
At 305 and 309 it was negative because it is going down
at 300 and 301 and 305 it was positive because its going up
at 302/303/304/306/309 she wasn't accelerating at all because it is staying the same
<h2>
Answer:</h2>
1.68 x 10⁻⁸Ωm
<h2>
Explanation:</h2>
The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;
R = ρL/A ------------------------(i)
Where;
A = πd² / 4 [where d = diameter of the wire]
From the question;
L = 6.90m
d = 2.15mm = 0.00215m
R = 0.0320Ω
First calculate the crossectional area (A) of the wire as follows;
A = πd² / 4
[Take π = 3.142]
d = 0.00215m
∴ A = 3.142 x (0.00215)² / 4
∴ A = 0.000003631m²
Now, substitute the values of A, L, and R into equation (i) as follows;
R = ρL/A
0.0320 = ρ x 6.90 / 0.000003631
0.0320 = 1900302.95 x ρ
Solve for ρ;
=> ρ = 0.0320 / 1900302.95
=> ρ = 1.68 x 10⁻⁸Ωm
Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm
100N to the left. Newton's 3rd law action and reaction
Answer:
The pressure is 
Explanation:
From the question we are told that
The gauge pressure at the mouth is 
The radius of the column is 
The speed of the liquid outside the body is 
The area of the column is 
The area inside the mouth 
Generally according to continuity equation

=> 
=> 
=> 
So

=> 
=> 
substituting values


Now the height of inside the mouth is 
Now the height of the column is 
Generally according to Bernoulli's equation
![p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%5Cfrac%7B1%7D%7B2%7D%20%20%5Crho%20v_2%5E2%20%2B%20h_2%20%5Crho%20g%20%2Bp_2%5D%20-%5B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%2A%20v_1%5E2%20%2B%20h_1%20%5Crho%20g%20%5D)
Now
which is the density of water
is the gauge pressure of the atmosphere which is zero
So
![p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%280.5%20%2A%201000%20%2A%20%283.1%29%5E2%29%20%2B%280.008%20%2A%201000%20%2A%209.8%29%20%2B%200%5D-)
