Determine the acceleration of the box of mass 140 kg, when a net force of 900 N acts on it.

In the physics of motion a force acting over a distance is?

lutik1710 [3]

A force over distance is work the unite is joules

8 0

3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what

pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector $F_{net}$ = vector Fe + vector Fn,

⇒ | $F_{net}$ | = $\sqrt{} F^{2} _{e } + F^{2}_n$

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ $F_{net}$ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ $F_{net}$ = √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0

2 years ago

If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

Given the following the details;

Energy = 40 Joules

Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

$Quantity \; of \; charge = \frac{Energy}{Electric \; potential}$

Substituting the values into the formula, we have;

$Quantity \; of \; charge = \frac{40}{8}$

Quantity of charge = 5 Coulombs

Therefore, the quantity of charge must be 5 Coulombs.

Find more information: brainly.com/question/21808222

8 0

3 years ago

Read 2 more answers

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C

Rus_ich [418]

Answer:

$1.50\ *10^{-6} }$

Explanation:

Given

e=100 N/C

M=0.15 g

$q=\ 22\ pC\\=\ 22\ *10^{-2}$

The ratio of the electric force on the bee to the bee's weight can be determined by the following formula

$\frac{fe}{M*9.81}$

$\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}$

$=\ 1.50\ *10^{-6}$

4 0

3 years ago

Determine the acceleration of the box of mass 140 kg, when a net force of 900 N acts on it.​

Determine the acceleration of the box of mass 140 kg, when a net force of 900 N acts on it.