Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
Answer:
v = √2G 
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1 / R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0
Answer
It should be A and C
Explanation:
because oxygen is number 8 in the periodic table of elements and has a atomic weight of 15.999 you use those numbers to figure out what is true between those.
The 8 for oxygen goes for the number of electrons and proton and to find neutrons u round the 15.999 up which now make it 16 and subtract it by the 8 now you have 8 protons, 8 neutrons, and 8 electrons
1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a
7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd
