All categories

3 years ago

A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is:

Physics

1 answer:

devlian [24]3 years ago

3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is option 5

Explanation:

From the question we are told that the positive particle is stationary hence the velocity is zero

Generally the magnetic force acting on the particle is mathematically represented as

$\vec F = Q \vec v \ \ X \ \vec B$

Here X represents a cross product

So since the velocity is zero

$\vec F = Q (0) \ \ X \ \vec B$

$\vec F = 0$

You might be interested in

A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength

Step2247 [10]

Answer:

Magnetic field, $B=4.16\times 10^{-5}\ T$

Explanation:

It is given that,

Velocity of proton, $v=3.5\times 10^7\ m/s$

Magnetic force, $F=1.65\times 10^{-16}\ N$

Charge of proton, $q=1.6\times 10^{-19}\ C$

We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :

$F=qvB\ sin\theta$

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}$

B = 0.0000416 T

$B=4.16\times 10^{-5}\ T$

Hence, this is the required solution.

4 0

3 years ago

How is rotational inertia defined? (1 point)

babunello [35]

Answer:

How is the rational inertia defined.

Answer. A

6 0

3 years ago

A crate (weight 292 N) is pushed 3.75 m up a 20.0° incline with a force of 125 N. The force of friction on the box is 25.0 N.

Firlakuza [10]

The acceleration of the box is $4.37\cdot 10^{-3} m/s^2$

Explanation:

The acceleration of the crate can be found by analyzing the forces acting along the direction parallel to the incline. We have three forces:

The force of push, F = 125 N, pushing up along the incline
The force of friction, $F_f = 25.0 N$ , acting down along the incline
The component of the weight parallel to the incline, $W sin \theta$ , also acting downward

Taking into account the direction of each force, the equation of motion along this direction is:

$F-F_f -W sin \theta = ma$

where

F = 125 N

$F_f = 25.0 N$

$W=292 N$ is the weight of the crate

$\theta=20.0^{\circ}$ is the angle of the incline

$m=\frac{W}{g}=\frac{292}{9.8}=29.8 kg$ is the mass of the box

a is the acceleration

And solving for a, we find:

$a=\frac{F-F_f - Wsin \theta}{m}=\frac{125-25-(292)(sin 20^{\circ})}{29.8}=4.37\cdot 10^{-3} m/s^2$

Learn more about forces and acceleration:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

7 0

4 years ago

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

6 0

3 years ago

A ball with an initial velocity of 5 m/s rolls off a horizontal table with a height of 1.0

Dmitry [639]

Answer:

Approximately $0.45\; \rm s$ , assuming that air resistance is negligible and that $g = 9.81\; \rm m\cdot s^{-2}$ .

Explanation:

The ball starts to fall the moment it rolls of the edge of the table.

Let $h_{0}$ denote the initially height of this ball.
Let $v_{0}$ denote the initial vertical velocity of this ball.

Assume that gravity is the only force acting on the ball during its fall (that is, there's no air resistance to slow the ball down.) The vertical acceleration of this ball during the fall would be constantly equal to $(-g)$ (negative because the ball is accelerating downwards.)

The following SUVAT equation would give the height $h$ of this ball at time $t$ :

$\displaystyle h = \frac{1}{2}\, (-g)\cdot t^{2} + v_{0} \cdot t + h_{0}$ .

Since the table is horizontal, the vertical velocity of this ball would be $0$ the moment it rolls of the edge. In other words: $v_{0} = 0\; \rm m\cdot s^{-1}$ .

The initial height of this ball when it rolls of the table is $h_{0} = 1.0\; \rm m$ , same as the height of the table.

Hence, the height $h$ of this ball at time $t$ would be:

$\begin{aligned}h &= \frac{1}{2}\, (-g)\cdot t^{2} + v_{0} \cdot t + h_{0} \\ &= \frac{1}{2}\, (-g) \cdot t^{2} + h_{0}\end{aligned}$ .

At $t = 0\; \rm s$ , the height of this ball would be $1.0\; \rm m$ . The ball would be on the ground by the time $h = 0\; \rm m$ , Set the right-hand side of this equation to $0$ and solve for the time $t$ at which the ball is on the ground:

$\displaystyle \frac{1}{2}\, (-g) \cdot t^{2} + h_{0} = 0$ .

$\displaystyle \frac{1}{2}\, (-g) \cdot t^{2} = - h_{0}$ .

$\displaystyle t = \sqrt{\frac{2\, h_{0}}{g}}$ .

Substitute in the values $h_{0} = 1.0\; \rm m$ and $g = 9.81\; \rm m\cdot s^{-2}$ :

$\begin{aligned} t &= \sqrt{\frac{2\, h_{0}}{g}} \\ &= \sqrt{\frac{1.0\; \rm m}{9.81\; \rm m\cdot s^{-2}}} \\ &\approx0.45\; \rm s\end{aligned}$ .

In other words, the ball would be in the air for approximately $0.45\; \rm s$ . (The initial horizontal velocity of this ball does not affect the duration of this fall.)

8 0

3 years ago