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3 years ago

How does the sum of forces help you

Physics

2 answers:

kirill115 [55]3 years ago

5 0

Answer:

idk

Explanation:

densk [106]3 years ago

3 0

Answer:

the net force is the resultant of all the forces; it is the result of adding all the forces together as vectors. For the situation of the three forces on the force board, the net force is the sum of force vectors A + B + C.

Explanation:

I hope this helps!!!

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Help will mark brain list

Hatshy [7]

A jagged line represents a resistor .

3 0

2 years ago

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

$h=u't+\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}\times 9.8\times (4.3)^2$

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

5 0

3 years ago

Explain how the potential energy of two charged particles depends on the distance between the charged particles and on the magni

maks197457 [2]

Answer:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges

Explanation:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges. Since the potential energy of two charged particles is directly proportional to the product of the two charges, its magnitude increases as the charges of the particles increases. For like charges, the potential energy is positive(the product of the two alike charges must be positive) and since potential energy is inversely proportional to the distance between the charges therefore it decreases as the particles get farther apart . For opposite charges, the potential energy is negative(the product of the two opposite charges must be negative) and since potential energy is inversely proportional to the distance between the two charges, it becomes more negative as the particles get closer together.

8 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

Please follow me guys <br>

Lilit [14]

Answer:

if i do give me brainliest ok ok

Explanation:

7 0

3 years ago