Answer:
Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .
Explanation:
Given :
Mass of brass, m₁ = 440 g
Temperature of brass, T₁ = 97° C
Mass of water, m₂ = 350 g
Temperature of water, T₂ = 23° C
Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹
Equilibrium temperature, T = 31° C
Let C₁ be the specific heat of brass.
Heat loss by brass = Heat gain by water
m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )
Substitute the suitable values in above equation.
440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)
C₁ = ![\frac{11704}{29040}](https://tex.z-dn.net/?f=%5Cfrac%7B11704%7D%7B29040%7D)
C₁ = 0.40 J g⁻¹ °C⁻¹
Answer:
Option D
Explanation:
<u><em>Given:</em></u>
Mass = m = 110 kg
Acceleration due to gravity = g = 9.8 m/s
<u><em>Required:</em></u>
Weight = W = ?
<u><em>Formula</em></u>
W = mg
<u><em>Solution:</em></u>
W = (110)(9.8)
W = 1078 N
We are given a series circuit with two light bulbs. In this case, the light bulbs act as resistors in series and the total resistance is:
![R_t=R_1+R_2](https://tex.z-dn.net/?f=R_t%3DR_1%2BR_2)
That is the sum of all the resistances in series in the circuit. To determine the voltage we can use Ohm's law:
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
Where "R" is the total resistance and "I" is the current in the circuit. Replacing we get:
PART A)
By Snell's law we know that
![n_1sin i = n_2 sin r](https://tex.z-dn.net/?f=n_1sin%20i%20%3D%20n_2%20sin%20r)
here we know that
![n_1 = 1.5](https://tex.z-dn.net/?f=n_1%20%3D%201.5)
![i = 25 degree](https://tex.z-dn.net/?f=i%20%3D%2025%20degree)
![n_2 = 1](https://tex.z-dn.net/?f=n_2%20%3D%201)
now from above equation we have
![1.5 sin25 = 1 sin r](https://tex.z-dn.net/?f=1.5%20sin25%20%3D%201%20sin%20r)
![r = 39.3 degree](https://tex.z-dn.net/?f=r%20%3D%2039.3%20degree)
so it will refract by angle 39.3 degree
PART B)
Here as we can see that image formed on the other side of lens
So it is a real and inverted image
Also we can see that size of image is lesser than the size of object here
Here we can use concave mirror to form same type of real and inverted image
PART C)
As per the mirror formula we know that
![\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7Bd_o%7D%20%3D%20%5Cfrac%7B1%7D%7Bf%7D)
![\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bd_i%7D%20%2B%20%5Cfrac%7B1%7D%7B60%7D%20%3D%20%5Cfrac%7B1%7D%7B20%7D)
![d_i = 30 cm](https://tex.z-dn.net/?f=d_i%20%3D%2030%20cm)
so image will form at 30 cm from mirror
it is virtual image and smaller in size