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3 years ago

How do the properties of individual elements change when they become part of a compound?

1 answer:

6 0

All I know is that the more proton and electrons an atom has the less metallic it becomes. But other than that I think it's pretty random

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An electric circuit can have no current when a switch is

Finger [1]

Answer:

Open

Explanation:

A switch is a part of a circuit where a connection can be made or broken. By convention, when the switch is "open", the connection is broken and current cannot pass. When the switch is "closed", the connection is complete and current can pass.

6 0

3 years ago

What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters

Setler79 [48]

Answer:A

Explanation:I TOOK THE Test

5 0

3 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

From the Coulomb's Law the force between the charges is given as:

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

Now force:

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

Now the acceleration on the raindrops due to the electrostatic force:

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

Read 2 more answers

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$