Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
Answer:
Althought it us possible to have more then one state, it's also possible to have only one state of matter
Explanation:
You can make solutions of only one state if matter, for example , it two liquids can be mixed to form a solution they are called miscible.
Answer:
<em> The elastic potential energy stored in the bungee cord = 20 J</em>
Explanation:
potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below
E = 1/2ke²................ Equation 1
Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension
<em>Given: K = 10 N/m, e = 2.00 m</em>
<em>Substituting these values into Equation 1</em>
<em>E = 1/2(10)(2)²</em>
<em>E = 5×4</em>
<em>E = 20 Joules.</em>
<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>
<em></em>
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows

![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)



T = 81.52 Deg C