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nikdorinn [45]
2 years ago
13

What are the two naturally occurring liquid elements?​

Physics
2 answers:
insens350 [35]2 years ago
7 0

Answer:

Explanation:

The two liquid elements are bromine and mercury. Many elements naturally occur as gases: hydrogen, nitrogen, oxygen, fluorine, chlorine, and all the noble gases (Group VIIIA elements). The remaining elements exist as solids.

kifflom [539]2 years ago
4 0

Answer:

bromine and mercury

Explanation:

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If are spaced closely together on the map,there is a drastic temperature change over the distance
AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0
3 years ago
Read 2 more answers
A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true
Fantom [35]

Answer:

(d) not enough info

Explanation:

because it doesn't specify where the strings are attached

if it was the two ends of the rod then T1 would be equal to T2

6 0
2 years ago
A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches the surface in 9.00 second
Natali5045456 [20]

The feather's vertical position y is determined by

y=\dfrac12g_{\text{moon}}t^2

We take the feather's starting position to be the origin, and the downward direction to be positive. Then

y=\dfrac12\left(1.63\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(9.00\,\mathrm s\right)^2=66.0\,\mathrm m

so the answer is D.

3 0
2 years ago
Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the
Leya [2.2K]

Answer: 1.298\times 10^{-5}\ N/m

Explanation:

Given

Current in the first wire I_1=2.2\ A

Current in the second wire I_2=5.9\ A

wires are 20\ cm apart

Force per unit length between the current-carrying wires is

\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}

Force exerted  by the wires is the same

Put the values

\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m

This force will be repulsive in nature as the current is flowing opposite

4 0
3 years ago
Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

4 0
3 years ago
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