Answer: 250n
Explanation:
The formula for gravitational force is: F = (gMm)/r^2
There are two factors at play here:
1) The mass of the planet 'M'
2) The radius 'r'
We can ignore the small M and the g, they are constants that do not alter the outcome of this question.
You can see that both M and r are double that of earth. So lets say earth has M=1 and r=1. Then, new planet would have M=2 and r=2. Let's sub these two sets into the equation:
Earth. F = M/r^2 = 1/1
New planet. F = M/r^2 = 2/4 = 1/2
So you can see that the force on the new planet is half of that felt on Earth.
The question tells us that the force on earth is 500n for this person, so then on the new planet it would be half! So, 250n!
Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
Answer:
1.3 x 10^(-2) atm/s
Explanation:
It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:
1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s
Answer: q = -52.5 μC
Explanation:
The complete question is given thus;
A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).
SOLVING:
from the given parameters we can solve this problem.
Given that the
Speed = 280 m/s
y = 70mm
B = -30 * 10⁻⁶T
Using the equation for magnetic field we have;
Β = μqv*r / 4πr²
making q (charge) the subject of formula we have that;
q = B * 4 *πr² / μqv*r
substituting the values gives us
q = (-0.3*10⁻⁶Tk * 4π * 0.07²) / (4π*10⁻⁷ * 280 ) = - [14.7 * 10⁻¹⁰k / 2.8 * 10⁻⁵ k ]
q = -52.5 μC
cheers i hope this helped !!!
