Answer:
C?
Explanation:
My best guess would be C as it's the only answer that gives a reason behind the statement.
Answer:
T = 74°C
Explanation:
Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g
Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium
Initial temperature of water = 100°C.
Initial temperature of aluminium = 29°C
When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.
Heat lost by water equal to the heat gained by aluminium pan.
Mw × Cw×(100 –T) = Ma × Ca × (T–29)
330×4.2×(100– T) = 890×0.9×(T–29)
1386(100 – T) = 801(T –29)
1386/801(100 – T) = T – 29
1.73(100 – T) = T – 29
173 –1.73T = T –29
173+29 = T + 1.73T
202 = 2.73T
T = 202/2.73
T = 74°C
Answer:
(A) Torque required is 21.205 N-m
(b) Wok done will be equal to 1199.1286 j
Explanation:
We have given moment of inertia 
Wheel deaccelerate from 135 rpm to 0 rpm
135 rpm = 
Time t = 8 sec
So angular speed 
and 
Angular acceleration is given by 
Torque is given by torque 
Work done to accelerate the vehicle is
Answer:
<em>The correct answer is (a) 312.5nm and (b) 125nm</em>
Explanation:
<em>The first step to take is to find The minimum thickness of the slick of the oil</em>
<em>Given that,</em>
<em>(a) tmin λ/2n </em>
<em>We substitute 750nm ( in air) for λ and 1.20 for n for the expression of minimum thickness t of the oil slick at that spot</em>
<em>thus,</em>
<em>tmin = (750nm)/2(1.2) = 312.5nm</em>
<em>The minimum thickness of the oil slick at that spot is =312.5nm</em>
<em>(B) we find the minimum thickness t </em>
<em>The minimum thickness of the oil slick at the spot will be calculated by,</em>
<em>tmin = λ/4n</em>
<em>we then 750nm ( in air) for λ and 1.50 for n in the expression for the minimum thickness of the slick of the oil.</em>
<em>tmin = (750nm)/4 (1.5) = 125nm</em>
<em>Therefore the minimum thickness t will now be = 125nm</em>
<em />
Work done is 0.442J
<u>Explanation:</u>
Given:
Spring constant, k = 33 N/m
Distance, x₁ = 0.15m
Additional distance, x₂ = 0.072 m
Total distance = 0.15 + 0.072 m
= 0.222 m
Work done, W = ?
We can calculate work done by the formula
On substituting the value we get:
![W = \frac{1}{2}k [(x_2)^2 - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%5B%28x_2%29%5E2%20%20-%20%28x_1%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20X%2033%5B%280.222%29%5E2%20-%20%280.15%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7DX%2033%20%5B%200.0493%20-%200.0225%5D%5C%5C%20%5C%5CW%20%3D%200.442%20J)
Therefore, work done is 0.442J
