Answer:
B. 1500 kg*m/s
Explanation:
Momentum p = m* v
In any type of collision, the total momentum is preserved!
The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.
p1 + p2 =
m1*v1 + m2*v2
m1 = me = 300 kg
v1 = 3 m/s
v2 = 2 m/s
Substitute the given numbers:
300*3 + 300+2
900 + 600
1500 kg*m/s, which is answer B.
Explanation:
It s given that,
Mass of a planet, ![M=3.7\times 10^{24}\ kg](https://tex.z-dn.net/?f=M%3D3.7%5Ctimes%2010%5E%7B24%7D%5C%20kg)
Radius of a planet, ![R=9.2\times 10^{6}\ m](https://tex.z-dn.net/?f=R%3D9.2%5Ctimes%2010%5E%7B6%7D%5C%20m)
(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :
![g=\dfrac{GM}{R^2}](https://tex.z-dn.net/?f=g%3D%5Cdfrac%7BGM%7D%7BR%5E2%7D)
![g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}](https://tex.z-dn.net/?f=g%3D%5Cdfrac%7B6.67%5Ctimes%2010%5E%7B-11%7D%5C%20Nm%5E2%2Fkg%5E2%5Ctimes%203.7%5Ctimes%2010%5E%7B24%7D%5C%20kg%7D%7B%289.2%5Ctimes%2010%5E%7B6%7D%5C%20m%29%5E2%7D)
![g=2.91\ m/s^2](https://tex.z-dn.net/?f=g%3D2.91%5C%20m%2Fs%5E2)
(2) The escape velocity is given by :
![v=\sqrt{\dfrac{2GM}{R}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B2GM%7D%7BR%7D%7D)
![v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%7B%5Cdfrac%7B2%5Ctimes%206.67%5Ctimes%2010%5E%7B-11%7D%5C%20Nm%5E2%2Fkg%5E2%5Ctimes%203.7%5Ctimes%2010%5E%7B24%7D%5C%20kg%7D%7B9.2%5Ctimes%2010%5E%7B6%7D%5C%20m%7D%7D)
v = 7324.61 m/s
Hence, this is the required solution.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
![M = \frac{q}{p}](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7Bq%7D%7Bp%7D)
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
![4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m](https://tex.z-dn.net/?f=4%20%3D%20%5Cfrac%7B5%5C%20m%7D%7Bp%7D%5C%5C%5C%5Cp%20%3D%20%5Cfrac%7B5%5C%20m%7D%7B4%7D%5C%5C%5C%5Cp%20%3D%201.25%5C%20m)
Now using thin lens formula:
![\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%3D%5Cfrac%7B1%7D%7Bp%7D%2B%5Cfrac%7B1%7D%7Bq%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7B1.25%5C%20m%7D%2B%5Cfrac%7B1%7D%7B5%5C%20m%7D%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bf%7D%20%3D%201%5C%20m%5E%7B-1%7D%5C%5C%5C%5C)
<u>f = 1 m</u>
The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.