Answer:
the force will increase by a factor 2.25
Explanation:
The gravitational force between the two stars is given by:

where
G is the gravitational constant
m1, m2 are the masses of the two stars
r is the distance between the stars
If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

So the new force will be

So, the force will be 2.25 times the previous value.
Answer:
Ok I'm not 100% on this one but, try 3 lifes sorry if u get it wrong D:
Explanation:
Explanation:
It is given that the number of electrons passing through the cross-sectional area in 1 s is
. Also, we know that charge on an electron is
, then negative charge crossing to the left per second is as follows.
I- =
I- = 0.544 A
As it is given that the number of protons crossing per second is
, as the charge on the proton is
, then positive charge crossing to the right per second is calculated as follows.
I+ =
I+ = 0.224 A
I = l I+ l + l I- l
So, I = 0.544 + 0.224
= 0.768 A
Thus, we can conclude that the current in given hydrogen discharge tube is 0.768 A.
Answer:
F = 32.28 N
Explanation:
For this exercise we must use the rotational equilibrium relation
Σ τ = 0
In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.
W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0
we assume that the magnitude of the forces applied by the hands is the same
F_left = F_right = F
W_bird L / 2 - 2 F 0.595 = 0
F =
we calculate
F = 0.560 9.8 14.0 /2.38
F = 32.28 N
4.3A.
The easiest way to solve this problem is find the equivalent resistance for parallel resistor 1/Req = 1/R1 + 1/R2 + 1/R3 in the three-branch parallel network with branches whose resistance are 8Ω.
1/Req = 1/8 Ω + 1/8 Ω + 1/8 Ω
1/Req = 3/8 Ω
Req = 8/3 Ω = 2.667Ω
Req = 2.7Ω
So, the equivalent circuit will be the 20.0V battery in series with a resistor 2.0Ω and the equivalent resistor 2.7Ω.
Using Ohm's Law to find the current provide by the 20.0V voltage source:
V = I*R ------> I = V/R
I = 20.0V/(2.0Ω + 2.7Ω)
I = 20.0V/4.7Ω
I = 4.3A