Answer:42.43m/s
Explanation:According to vf=vi+at, we can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s
The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge
C is placed which is the origin.
Let B be the point where the charge
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law,
,
= 



k is the Coulomb's law constant.
On substituting the values into the above equation, we get,

Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
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Answer:
The Moon's distance from the Earth varies during its orbit
Explanation:
The correct statement is ,The Moon's distance from the Earth varies during its orbit.
Important point regarding moon:
1 .Moon is a natural satellite of the earth.
2. Moon is the fifth largest satellite in solar system.
3.Second densest satellite in solar system.
4.Moon rotates about earth.
5.Moon is an astronomical body .
<h2>
Answer: 0.17</h2>
Explanation:
The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":
(1)
Where:
is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 
is the Stefan-Boltzmann's constant.
is the Surface area of the body
is the effective temperature of the body (its surface absolute temperature) in Kelvin.
However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:
(2)
Where
is the body's emissivity
(the value we want to find)
Isolating
from (2):
(3)
Solving:
(4)
Finally:
(5) This is the body's emissivity
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.