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3 years ago

Which Hand Is Negativity Charged? (Picture included)

Physics

2 answers:

Vera_Pavlovna [14]3 years ago

6 0

B is negatively charged.

The electrons outnumber the protons, showing that its charge is negative.

However, on the rest of the hands, the number of electrons are either equal to (making the charge neutral) or less than (making the charge positive) the number of protons.

Hope this helped!

pantera1 [17]3 years ago

3 0

B. The number of electrons is greater. Causing the hand to become negatively charged.

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How are elastic and inelastic collisions different?

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Describe how reactivity changes as you go down Group 1A.

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Explanation:

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Suppose a small quantity of radon gas, which has a half-life of 3.8 days, is accidentally released into the air in a laboratory.

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Answer:

1 day

Explanation:

Let the safe level = x

The current level = x + 0.2x = 1.2 x

Thus,

Half life = 3.8 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{3.8}\ days^{-1}$

The rate constant, k = 0.1824 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}$ = x / 1.2 x = 0.8333

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

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4 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

How are longitudinal and transverse waves alike? How are they different?

larisa86 [58]

Answer:

Longitudinal and transverse waves have many similarities and differences.

Explanation:

Similarities:

Mechanical waves can be transverse and longitudinal waves.

Transverse and longitudinal waves both have wavelengths and frequencies.

They both have amplitudes

Both waves can travel through a medium or not, but it depends on whether is an electromagnetic or a mechanical wave.

Differences:

Electromagnetic waves can only be transverse.

The particles of the medium in a longitudinal wave move parallel to the direction (motion) of a wave. It is in this back and forth motion.

The particles of the medium in a transverse wave move perpendicular to the direction (motion) of a wave. This means that there would be right angles showing that they are perpendicular.

Longitudinal waves have rarefactions and compressions.

These rarefactions and compressions are used to measure the wavelength of a wave. For instance, a wavelength in a longitudinal wave is measured from rarefaction to rarefaction

Transverse waves have troughs and crests.

Amplitude in a transverse wave is measured from the midline to the crest of trough.

Amplitude in a longitudinal wave is measured based on how closely packed the particles of the medium are

I hope this helps

4 0

3 years ago