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3 years ago

Does displacement must be straight and must include direction

1 answer:

7 0

Displacement is the total space between point a and b. For example, If I walk 3 times North, 6 times East, and 7 times North, what is the displacement? The answer is 10 because when I walked East there was on object blocking my way.

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PLEASE HELP ME What does a moving charge experience when it is near a magnetic field? static friction gravity a stationary charg

attashe74 [19]

A moving charge experiences a force as it moves close to a magnetic field present in a region.

Explanation:

A charged particle when moves near a magnetic field is forced to move in a circular path. If there is a straight moving charged particle enters the region of the magnetic field, it's path is changed from straight path to the curved path.

The motion of the charged path inside or near the magnetic field will be a circular path and this circular path is due to the force experienced by the charged particle in the magnetic field.

The force experienced by the a charged particle as it moves near the magnetic field is termed as the Lorentz force. The expression for the Lorentz force experienced by the charged particle in the magnetic field is given by:

$\boxed{F_m=Q(\overrightarrow{v}\times\overrightarrow{B})}$

Here, $F_m$ is the magnetic force, $Q$ is the amount of charge, $\overrightarrow{v}$ is the velocity vector and $\overrightarrow{B}$ is the magnetic field intensity in the region.

Thus, A moving charge experiences a force as it moves close to a magnetic field present in a region.

Learn More:

1. Which one is true for the electromagnetic spectrum brainly.com/question/1619496

2. What is the current in resistor R2 brainly.com/question/3051098

3. Average current denisty of a linear wire brainly.com/question/10597501

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetism

Keywords:

moving, charge, straight, circular, path, magnetic, field, amount, force, velocity, intensity, Lorentz, expression, stationary.

6 0

2 years ago

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

Horizontal Launch

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago

If the speed of an object does NOT change, the object is traveling at a

Elanso [62]

Answer:

If the speed does not change at all, the object would be moving at a constant speed.

5 0

3 years ago

Equipotentials are lines along which Select one: a. the electric field is constant in magnitude and direction. b. the electric c

soldi70 [24.7K]

Answer:

option D is correct

Explanation:

It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along an equipotential, since ΔV = 0. Thus the work is :

W = −ΔPE = −qΔV = 0.

Work is zero if force is perpendicular to motion. Force is in the same direction as E, so that motion along an equipotential must be perpendicular to E. More precisely, work is related to the electric field by:

W = Fd cos θ = qEd cos θ = 0.

- The change in kinetic energy Δ K.E by conservation should be:

Δ K.E = W

Since, W = 0:

Δ K.E = 0

- If change in kinetic energy is zero it means that charge moves at a constant speed. Hence, option D is correct.

4 0

3 years ago

A simple pendulum is swinging back and forth through a small angle, its motion repeating every 1.15 is. How much longer should t

max2010maxim [7]

Answer:

ΔL=0.14m

Explanation:

The period of oscillations of simple pendulum is

$T=2\pi \sqrt{\frac{L}{g} } \\$

Here L is length of pendulum and g is acceleration due to gravity

So the length of pendulum is:

$L=\frac{T^{2}g }{4 \pi^{2} }\\$

So increase in length is:

ΔL=L₂-L₁

$=\frac{T_{2}^{2}g }{4 \pi^{2} }-\frac{T_{1}^{2}g }{4 \pi^{2} }\\=\frac{g}{4 \pi^{2} }[T_2^{2} -T_1^{2} ]\\ =\frac{9.8}{4 \pi^{2}}[(1.15+0.22)^{2}-(1.15)^{2} ] \\=0.14m$

ΔL=0.14m

3 0

3 years ago