Answer:
The quantity of heat given to body depends on: (i) The mass of the body, (ii) The rise (or fall) in the temperature of the body, and (iii) Nature of the material of the body.
Explanation:
Hope this helps
have a great day
~Zero
Answer:
Option c
Explanation:
Soil tilth refers to the physical condition of the soil, particularly with regards to the suitability of the soil for the growth of crop.
The determinants of the tilth in the soil incorporates the arrangement and steadiness of aggregated particles of the soil, pace of water penetration, level of air circulation, dampness content and seepage.
Thus option C follows the definition of the soil tilth.
Answer:
q = -2 m and q = -0.5 m
Explanation:
For this exercise we must use the equation of the optical constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image, respectively
Let's start with the far vision point, in this case the power of the lens is
P = -0.5D
power is defined as the inverse of the focal length in meter
f = 1 / D
f = -1 / 0.5
f = - 2m
the object for the far vision point is at infinity p = infinity
1 / f = 1 / p + i / q
1 / q = 1 / f - 1 / p
1 / q = -1/2 - 1 / ∞
q = -2 m
The sign indicates that the image is on the same side as the object
Now let's lock the near view point
D = +2.00 D
f = 1 / D
f = 0.5m
the near mink point is p = 25 cm = 0.25 m
1 / f = 1 / p + 1 / q
1 / q = 1 / f - 1 / p
1 / q = 1 / 0.5 - 1 / 0.25
1 / q = -2
q = -0.5 m
the sign indicates that the image is on the same side as the object in front of the lens
To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:
Where,
V = the linear speed
r = Radius
Angular speed
The angular speed is given by
Replacing at our first equation we have that the centripetal acceleration would be
To transform it into multiples of the earth's gravity which is given as 
the equivalent of 1g.
PART B) Now the linear speed would be subject to:
Therefore the linear speed of a point on its edge is 51.05m/s
Answer:
The electric field at x = 3L is 166.67 N/C
Solution:
As per the question:
The uniform line charge density on the x-axis for x, 0< x< L is 
Total charge, Q = 7 nC = 
At x = 2L,
Electric field, 
Coulomb constant, K = 
Now, we know that:
Also the line charge density:
Thus
Q = 
Now, for small element:
Integrating both the sides from x = L to x = 2L
![\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]](https://tex.z-dn.net/?f=%5Cvec%7BE_%7B2L%7D%7D%20%3D%20K%5Clambda%5B%5Cfrac%7B-%201%7D%7Bx%7D%5D_%7BL%7D%5E%7B2L%7D%5D%20%3D%20K%5Cfrac%7BQ%7D%7BL%7D%5Bfrac%7B1%7D%7B2L%7D%5D)
![\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE_%7B2L%7D%7D%20%3D%20%289%5Ctimes%2010%5E%7B9%7D%29%5Cfrac%7B7%5Ctimes%2010%5E%7B-%209%7D%7D%7BL%7D%5Bfrac%7B1%7D%7B2L%7D%5D%20%3D%20%5Cfrac%7B63%7D%7BL%5E%7B2%7D%7D)
Similarly,
For the field in between the range 2L< x < 3L:
![\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20K%5Clambda%5B%5Cfrac%7B-%201%7D%7Bx%7D%5D_%7B2L%7D%5E%7B3L%7D%5D%20%3D%20K%5Cfrac%7BQ%7D%7BL%7D%5Bfrac%7B1%7D%7B6L%7D%5D)
![\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%289%5Ctimes%2010%5E%7B9%7D%29%5Cfrac%7B7%5Ctimes%2010%5E%7B-%209%7D%7D%7BL%7D%5Bfrac%7B1%7D%7B6L%7D%5D%20%3D%20%5Cfrac%7B63%7D%7B6L%5E%7B2%7D%7D)
Now,
If at x = 2L,
Then at x = 3L:
