Physics: a particle has a constant velocity of 10 m/s and a constant acceleration. what is it’s velocity 1s later? 5s later?

Since the particle has the constant velocity Therefore it doesn t change with time. Therefore velocity...

a particle has a constant velocity of 10 m/s, №15194774, 04.04.2020 07:57

Physics

a particle with an initial velocity of 50 m slows at a constant acceleration to 20 ms-1 over a distance of 105 m. how long does it take for the particle to slow down? (a) 2 s (c) 3 s (b)4 s (d)5 s

Step-by-step answer

P Answered by PhD

Time taken, t = 3 s

Explanation:

It is given that,

Initial velocity of the particle, u = 50 m/s

Final velocity, v = 20 m/s

Distance covered, s = 105 m

Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :

v^2-u^2=2as

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}$

$a=-10\ m/s^2$

So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}$

t = 3 s

So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.

Physics

a particle with an initial velocity of 50 m slows at a constant acceleration to 20 ms-1 over a distance of 105 m. how long does it take for the particle to slow down? (a) 2 s (c) 3 s (b)4 s (d)5 s

Step-by-step answer

P Answered by PhD

Time taken, t = 3 s

Explanation:

It is given that,

Initial velocity of the particle, u = 50 m/s

Final velocity, v = 20 m/s

Distance covered, s = 105 m

Firstly we will find the acceleration of the particle. It can be calculated using third equation of motion as :

v^2-u^2=2as

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(20\ m/s)^2-(50\ m/s)^2}{2\times 105\ m}$

$a=-10\ m/s^2$

So, the particle is decelerating at the rate of 10 m/s². Let t is the time taken for the particle to slow down. Using first equation of motion as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{20\ m/s-50\ m/s}{-10\ m/s^2}$

t = 3 s

So, the time taken for the particle to slow down is 3 s. Hence, this is the required solution.

Physics

A charged particle with charge of 2 (uC) and mass 10-20 (kg) is traveling with velocity of 108 (m/s) in space. The charge reaches to a region in space with magnetic field of 0.05 (T) and experience a force of 8 (N) exerted by the magnetic field. A- What is the angle between velocity of particle and magnetic field direction? B- What is acceleration of charged particle while experiencing the force?

Step-by-step answer

P Answered by Master

Explanation:

A ) Let the angle be θ between magnetic field and velocity of charged particle

Force created on charged particle F

= Bqv sinθ, B is magnetic field , q is charge , v is velocity of charged particle

F = .05 x 2 x 10⁻⁶ x 10⁸ x sinθ

8 = 10 sinθ

sinθ = .8

θ = 53°.

B )

acceleration = force / mass

= 8 / 10⁻²⁰

= 8 x 10²⁰ m / s²

Physics

An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045T. (An alpha particle has a charge of charge of +3.2 x 10-19 C and a mass 6.6 x 10-27 kg) The angle between the particle s direction of motion and the magnetic field is 52°. What happens to the speed? The answer is that the speed remains the same and I want to know why and how would we know if the speed changes ( there are two questions before this about the force and acceleration force is 6.2x10^-18 and acceleration is 9.5x10^8 )

Step-by-step answer

P Answered by PhD

the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Explanation:

The magnetic force is given by the expression

F = q v x B

The bold are vectors, where v is the velocity and B is the magnetic field, the product is the cross product whose result is a vector perpendicular to the two vectors (v and B)

From the above, the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Even when the change in direction is real and is caused by a centripetal force

For there to be a change in the velocity modulus there must be a force parallel to the velocity direction, generally a force in electrical

Physics

Aparticle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field b⃗ =(− 0.130t) k^. at a particular instant of time the velocity of the particle is v⃗ =(1.20 × 106 m/s) (−3i^ + 4j^+12 k^) and the force f⃗ on the particle has a magnitude of 2.30 n . (a) determine the charge q. (b) determine the acceleration a⃗ of the particle. (c) find the x, y and z-component of the acceleration

Step-by-step answer

P Answered by PhD

q = 2,95 10-6 C

Explanation:

The magnetic force on a particle is described by the equation

F = q v x B

Where bold indicate vectors

Let's make the vector product

vxB = $\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]$

v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

vx B = 1.20 106 [0.52 i ^ - 0.39j ^]

As they give us the force module, let's use Pythagoras' theorem,

|v xB | =1.20 10⁶ √( 0.52² + 0.39²)

|v x B| = 1.20 10⁶ 0.65

v xB = 0.78 10⁶

Let's replace and calculate

2.30 = q 0.78 10⁶

q = 2.3 / 0.78 106

q = 2,95 10-6 C

Physics

Aparticle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field b⃗ =(− 0.130t) k^. at a particular instant of time the velocity of the particle is v⃗ =(1.20 × 106 m/s) (−3i^ + 4j^+12 k^) and the force f⃗ on the particle has a magnitude of 2.30 n . (a) determine the charge q. (b) determine the acceleration a⃗ of the particle. (c) find the x, y and z-component of the acceleration

Step-by-step answer

P Answered by PhD

q = 2,95 10-6 C

Explanation:

The magnetic force on a particle is described by the equation

F = q v x B

Where bold indicate vectors

Let's make the vector product

vxB = $\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]$

v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

vx B = 1.20 106 [0.52 i ^ - 0.39j ^]

As they give us the force module, let's use Pythagoras' theorem,

|v xB | =1.20 10⁶ √( 0.52² + 0.39²)

|v x B| = 1.20 10⁶ 0.65

v xB = 0.78 10⁶

Let's replace and calculate

2.30 = q 0.78 10⁶

q = 2.3 / 0.78 106

q = 2,95 10-6 C

Physics

A charged particle with charge of 2 (uC) and mass 10-20 (kg) is traveling with velocity of 108 (m/s) in space. The charge reaches to a region in space with magnetic field of 0.05 (T) and experience a force of 8 (N) exerted by the magnetic field. A- What is the angle between velocity of particle and magnetic field direction? B- What is acceleration of charged particle while experiencing the force?

Step-by-step answer

P Answered by Master

Explanation:

A ) Let the angle be θ between magnetic field and velocity of charged particle

Force created on charged particle F

= Bqv sinθ, B is magnetic field , q is charge , v is velocity of charged particle

F = .05 x 2 x 10⁻⁶ x 10⁸ x sinθ

8 = 10 sinθ

sinθ = .8

θ = 53°.

B )

acceleration = force / mass

= 8 / 10⁻²⁰

= 8 x 10²⁰ m / s²

Physics

An alpha particle travels at a velocity of magnitude 550 m/s through a uniform magnetic field of magnitude 0.045T. (An alpha particle has a charge of charge of +3.2 x 10-19 C and a mass 6.6 x 10-27 kg) The angle between the particle s direction of motion and the magnetic field is 52°. What happens to the speed? The answer is that the speed remains the same and I want to know why and how would we know if the speed changes ( there are two questions before this about the force and acceleration force is 6.2x10^-18 and acceleration is 9.5x10^8 )

Step-by-step answer

P Answered by PhD

the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Explanation:

The magnetic force is given by the expression

F = q v x B

The bold are vectors, where v is the velocity and B is the magnetic field, the product is the cross product whose result is a vector perpendicular to the two vectors (v and B)

From the above, the force is perpendicular to the speed, it is a type of force that changes the direction of the speed, as in the uniform circular motion te, but does not change its modulus.

Even when the change in direction is real and is caused by a centripetal force

For there to be a change in the velocity modulus there must be a force parallel to the velocity direction, generally a force in electrical

Physics

An alpha particle travels at a velocity of magnitude 520 m/s through a uniform magnetic field of magnitude 0.034 T. (An alpha particle has a charge of charge of +3.2 × 10-19 C and a mass 6.6 × 10-27 kg) The angle between the particle s direction of motion and the magnetic field is 43°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) Does the speed of the particle increase, decrease, or remain the same?

Step-by-step answer

P Answered by Specialist

(a) Magnetic force $F=38.584\times 10^{-19}N$