Answer:
The coordinates of the point is (0,0.55).
Explanation:
Given that,
First charge 
at origin
Second charge 
Second charge at point P = (0,1)
We assume that,
The net electric field between the charges is zero at mid point.
Using formula of electric field
Hence, The coordinates of the point is (0,0.55).
We don't know the change in velocity, so can't answer.
Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º
Answer:In a series circuit, adding more resistors increases total resistance and thus lowers current. But the opposite is true in a parallel circuit because adding more resistors in parallel creates more choices and lowers total resistance. If the same battery is connected to the resistors, current will increase.
Explanation:
<span>Density is 3.4x10^18 kg/m^3
Dime weighs 1.5x10^12 pounds
The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so
4/3 pi 1.9x10^3
= 4/3 pi 6.859x10^3 m^3
= 2.873x10^10 m^3
Now divide the mass by the volume
9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3
Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3
Now to figure out how much the dime weighs, just multiply by the volume of the dime.
3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg
And to convert from kg to lbs, multiply by 2.20462, so
6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
