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3 years ago

A charged particle is surrounded by an electric field and a magnetic field

Physics

2 answers:

zlopas [31]3 years ago

7 0

Answer:

yeah

Explanation:

electric fields help the charged particles interact

and isn't magnetic field the same as electric field.

mihalych1998 [28]3 years ago

7 0

A charged particle is surrounded by an electric field (E field), but no magnetic field. When the particle moves, the changing electric field induces a magnetic field: this is one of the Maxwell Equations (Ampère’s Law)

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how many years would it take to reach the star from earth, as measured by observers on the spacecraft

Assoli18 [71]

In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.

Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.

Learn more about Light year here-

brainly.com/question/1302132

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3 0

1 year ago

An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time

blagie [28]

Answer:

100m

Explanation:

$s = ut + \frac{1}{2} a {t}^{2}$

u=0;t=10sec;a=2m/s²

$s = 0(10) + \frac{1}{2}(2 \times {10}^{2} )$

s=10²;100m

7 0

3 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago

A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav

alexdok [17]

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

t = 3.5 sec

5 0

3 years ago

A person hits a 45-g golf ball. The ball comes down on a tree root and bounces

Verizon [17]

Answer:

Maximum height, h = 10 m

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

The ball comes down on a tree root and bounces straight up with an initial speed of 14.0 m/s.

We need to find the height the ball will rise after the bounce. It is based on the conservation of energy such that,

$\dfrac{1}{2}mv^2=mgh$

h is maximum height raised by the ball

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m$

So, the ball will raised to a height of 10 meters.

5 0

3 years ago