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3 years ago

How do stationary charged particles interact at a distance with their surroundings?

Physics

1 answer:

quester [9]3 years ago

7 0

Answer:

The answer is: by way of their electric fields.

Explanation:

When it comes to Physics, a<u> "stationary charge" </u><u>is a particle that is moving</u>. It is not acted upon by a magnetic force because it does not interact with "static magnetic fields."

Although it cannot create magnetic force fields, it can interact at a distance with their surroundings by way of their electric fields. This electric field is a result of imbalance in the electric charges and it doesn't affect magnets. However, this can, somehow, be extended and felt in their environment.

Thus, this explains the answer.

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s

yaroslaw [1]

Answer : The value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = ?

t = time = 17s

$[A_t]$ = final concentration = 0.0981 M

$[A_o]$ = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

$k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}$

$k=0.51M^{-1}s^{-1}$

Therefore, the value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

6 0

3 years ago

The internal energy of material is determined by

Mnenie [13.5K]

The combined amount of kinetic and potential energy of its molecules

6 0

4 years ago

Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par

kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0

4 years ago

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is

Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_{f}$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_{f}$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_{f}$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

5 0

3 years ago