Answer:
5.63 gram
Explanation:
As per the equation
Q = Lm
where Q is the amount of heat needed to convert water in to vapour
and L is the latent heat f vaporization.
m is the mass converted, hence putting all the values.
m =Q/L
= 12725400/ 2260000
= 5.63 gram
the common attributes are positive and negatively charged
The acceleration of the car,
![a = \frac{v-u}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv-u%7D%7Bt%7D)
Here, v is final velocity, u is initial velocity and t is time taken by the car.
Given
,
and ![t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h](https://tex.z-dn.net/?f=t%20%3D%203.0%20s%20%3D%203.0%20%5Ctimes%20%5Cfrac%7B1%20%5C%20h%7D%7B3600%7D%20%3D%208.3%20%5Ctimes%2010%5E%7B-4%7D%20h)
Therefore, from above equation
.
Here, negative sign shows deceleration of a car.
Thus the the magnitude of car acceleration is
.
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance
![\rho(r) = ar^2 - br^3](https://tex.z-dn.net/?f=%5Crho%28r%29%20%3D%20ar%5E2%20-%20br%5E3)
r is the radius of the spherical shell
dr is the thickness
volume of shell
![dV = 4 \pi r^2 dr](https://tex.z-dn.net/?f=dV%20%3D%204%20%5Cpi%20r%5E2%20dr)
mass of shell
![dM = \rho(r)dV](https://tex.z-dn.net/?f=dM%20%3D%20%5Crho%28r%29dV)
![\rho = \rho_0 - br](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Crho_0%20-%20br)
now,
![dM = (\rho_0 - br)(4 \pi r^2)dr](https://tex.z-dn.net/?f=dM%20%3D%20%28%5Crho_0%20-%20br%29%284%20%5Cpi%20r%5E2%29dr)
integrating both side
![M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr](https://tex.z-dn.net/?f=M%20%3D%20%5Cint_0%5E%7BR%7D%20%28%5Crho_0%20-%20br%29%284%20%5Cpi%20r%5E2%29dr)
![M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})](https://tex.z-dn.net/?f=M%20%3D%20%5Cdfrac%7B4%5Cpi%7D%7B3%7DR%5E3%5Crho_0%20-%20%5Cpi%20R%5E4%28%5Cdfrac%7B%5Crho_0-%5Crho%7D%7BR%7D%29)
![M = \pi R^3(\dfrac{\rho_0}{3}+\rho)](https://tex.z-dn.net/?f=M%20%3D%20%5Cpi%20R%5E3%28%5Cdfrac%7B%5Crho_0%7D%7B3%7D%2B%5Crho%29)
we know,
![a = \dfrac{GM}{R^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cdfrac%7BGM%7D%7BR%5E2%7D)
![a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cdfrac%7BG%28%20%5Cpi%20R%5E3%28%5Cdfrac%7B%5Crho_0%7D%7B3%7D%2B%5Crho%29%29%7D%7BR%5E2%7D)
![a =\pi RG(\dfrac{\rho_0}{3}+\rho)](https://tex.z-dn.net/?f=a%20%3D%5Cpi%20RG%28%5Cdfrac%7B%5Crho_0%7D%7B3%7D%2B%5Crho%29)
![a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)](https://tex.z-dn.net/?f=a%20%3D%5Cpi%20%286.674%5Ctimes%2010%5E%7B-11%7D%5Ctimes%206.38%20%5Ctimes%2010%5E6%29%28%5Cdfrac%7B1.60%5Ctimes%2010%5E4%7D%7B3%7D%2B2.1%5Ctimes%2010%5E3%29)
a = 9.94 m/s²
Answer:
u = - 20 cm
m =![\frac{1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D)
Given:
Radius of curvature, R = 10 cm
image distance, v = 4 cm
Solution:
Focal length of the convex mirror, f:
f = ![\frac{R}{2} = \frac{10}{2} = 5 cm](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B2%7D%20%3D%20%5Cfrac%7B10%7D%7B2%7D%20%3D%205%20cm)
Using Lens' maker formula:
![\frac{1}{f} = \frac{1}{u} + \frac{1}{v}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bu%7D%20%2B%20%5Cfrac%7B1%7D%7Bv%7D)
Substitute the given values in the above formula:
![\frac{1}{5} = \frac{1}{u} + \frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%20%3D%20%5Cfrac%7B1%7D%7Bu%7D%20%2B%20%5Cfrac%7B1%7D%7B4%7D)
![\frac{1}{u} = \frac{1}{5} - \frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bu%7D%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20-%20%5Cfrac%7B1%7D%7B4%7D)
u = - 20 cm
where
u = object distance
Now, magnification is the ratio of image distance to the object distance:
magnification, m =![\frac{|v|}{|u|}](https://tex.z-dn.net/?f=%5Cfrac%7B%7Cv%7C%7D%7B%7Cu%7C%7D)
magnification, m =![\frac{|4|}{|-20|}](https://tex.z-dn.net/?f=%5Cfrac%7B%7C4%7C%7D%7B%7C-20%7C%7D)
m =![\frac{4}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B20%7D)
m =![\frac{1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D)