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2 years ago

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What is the current when a typical static charge of 0.234~\mu\text{c}0.234 μc moves from your finger to a metal doorknob in 0.59

5~\mu\text{s}0.595 μs?

1 answer:

Black_prince [1.1K]2 years ago

4 0

The current is defined as the quantity of charge Q that passes through a certain location in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Using the data of the problem, we find:
$I= \frac{Q}{\Delta t} = \frac{0.234 \mu C}{0.595 \mu s}=\frac{0.234\cdot 10^{-6} C}{0.595 \cdot 10^{-6} s}=0.39 A$

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Can you help with this question please

Iteru [2.4K]

Answer:

First answer to the first question is Two people pulling a rope with the same force in a opposite direction. The other one would be 2.72N

Explanation:

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2 years ago

Q.01 When charging a secondary cell, energy is stored within a dielectric material using an electric field. True or False

Nitella [24]

True, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

<h3>Relationship between dielectric material and electric field</h3>

The electric field in a capacitor separates the negative and positive charges in the dielectric material, this causes an attractive force between each plate and the dielectric.

The dielectric material can store electric energy due to its polarization in the presence of external electric field, which causes the positive charge to store on one electrode and negative charge on the other.

Thus, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

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4 0

2 years ago

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

2 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

2 years ago

What is the acceleration of an 24 kg object that applies a force of 130N?

WITCHER [35]

Answer:

To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.

Explanation:

Given ,

F = 130N

M = 24kg

A = ?

F = m× a

then ,

130N = 24kg ×a

a = 130/24 = 5 m/s.

6 0

1 year ago