In which direction does the electric field point at a position directly ((north))of a

Two boxes are 8 cm apart. Which of the following should Janet do to decrease the gravitational force between the boxes?

OlgaM077 [116]

Answer:

the answer is 2.

Explanation:

4 0

2 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

What kind of energy is in a rock at the edge of a cliff.

soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

8 0

2 years ago

The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is

mestny [16]

Answer:

3924 Pa

Explanation:

Volume of cylinder = 2 L = 0.002 m^3 (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = $\pi r^{2}$

where $\pi$ = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x $0.04^{2}$ = 0.005 m^2

height h of the water in this cylinder = volume/area

h = 0.002/0.005 = 0.4 m

pressure at the bottom of the cylinder due to the height of water = pgh

where

p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = 3924 Pa

3 0

3 years ago

A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the

lina2011 [118]

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

8 0

3 years ago