Waves interact with ___ and other __

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

5 0

2 years ago

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(c) Magnitude of the force that keeps you go around at this acceleration

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

3 years ago

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

4 0

3 years ago

Compare the collision between two baseballs and a catcher's mitt.

Nookie1986 [14]

The applied force is different for the two cases

The case A with a greater force involves the greatest momentum change

The case A involves the greatest force.

<h3>What is collision?</h3>

This is the head-on impact between two object moving in opposite or same direction.

The initial momentum of the two ball is the same.

P = mv

where;

m is the mass of each
v is the initial velocity of each ball

Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.

Thus, the final momentum of each ball will be different

The impulse experienced by each ball is different since impulse is the change in momentum of the balls.

J = ΔP

The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.

Thus, we can conclude the following;

The applied force is different for the two cases
The case A with a greater force involves the greatest momentum change
The case A involves the greatest force.

Learn more about impulse here: brainly.com/question/25700778

3 0

3 years ago

Picture showing four cards in a row with labels 'First animal appears on land', 'First life form appears in the sea', 'First roc

sasho [114]

The answer would be 'First rocks formed on Earth'.

3 0

3 years ago

Waves interact with ___ and other ___.

Waves interact with _ and other _.