Answer:
B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.
Explanation:
Given that Planets A and B have the same size, mass.
Let the masses of the planets A and B are 
and 
respectively.
As masses are equal, so 
.
Similarly, let the radii of the planets A and B are 
and 
respectively.
As radii are equal, so 
.
Let my mass is m.
As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.
So, my weight on planet A, 
my weight of planet B, 
By using equations (i) and (ii),
.
So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.
Hence, option (B) is correct.
From the gravity acceleration theorem due to a celestial body or planet, we have that the Force is given as
Where,
F = Strength
G = Universal acceleration constant
M = Mass of the planet
m = body mass
r = Distance between centers of gravity
The acceleration by gravity would be given under the relationship
Here the acceleration is independent of the mass of the body m. This is because the force itself depended on the mass of the object.
On the other hand, the acceleration of Newton's second law states that
Where the acceleration is inversely proportional to the mass but the Force does not depend explicitly on the mass of the object (Like the other case) and therefore the term of the mass must not necessarily be canceled but instead, considered.
In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W
Answer:
The sum of positive and negative charges in a unit of Al2O3 equals zero.
Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.
Al203 has 2 Al atoms and 3 O atoms.
Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)
= 2(3) + 3(-2)
= 6 - 6
= 0
Explanation:
Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.
Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.
