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stellarik [79]
3 years ago
11

Helium is an odorless, colorless gas at room temperature. It is less dense than air. That's why helium balloons float. It is a n

on-reactive gas, a member of the Noble gas family. Its melting point is -272.2 oC and its boiling point is -268.93 o A chemical property of Helium is? A) it’s boiling point of -268.93 oC B) the fact that it’s less dense than air C) It’s lack of color (colorless) D) It’s non-reactivity
Physics
2 answers:
harina [27]3 years ago
7 0
D I would think helium is not reactive to color or a boiling point
Alex777 [14]3 years ago
6 0

Answer:

D. It's non-reactivity

Explanation:

reactiveness is chemical

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How many significant figures does the value 0.080 have? 1 2 3 4
lara [203]
2. The zeros in front do not matter
3 0
3 years ago
List and describe the top 3 qualities of an<br> effective leader?
s2008m [1.1K]

BEING HONEST , OPTIMISTIC,DISCIPLINED, PUNCTUAL ,TRUTHFUL AND GOOD ORATOR(SPEAKER) , AND HOPEFUL ARE OTHER BASIC QUALITIES AN EFFECTIVE LEADER MUST POSSESS. ALSO,HE MUST NEVER DIFFERENCE BETWEEN ANY OF HIS FOLLOWERS OR DO PARTIALLY AT ANY TIME . HE SHOULD TREAT EVERY ONE EQUALLY.

Explanation:

7 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0
3 years ago
Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10
USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

y = \frac{1}{2} * g * t^{2}

where y = height = 40 ft.

As all the parameters are given in SI units, it is  advisable to convert this value to m, as follows:

y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m

Now, we can solve for t, as follows:

t = \sqrt{\frac{2*y}{g}} =  \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

y = v0*t +\frac{1}{2} * g * t^{2}

This leaves us a quadratic equation on t, as follows:

t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0
3 years ago
Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in a
den301095 [7]

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

Substituting the values of the variables into the equation, we have

I =  P/4πr²

I =  150.0 W/4π(6 m)²

I =  150.0 W/4π(36 m²)

I =  150.0 W/452.39 m²

I =  0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E

= √2 × 11.2 V/m

= 15.81 V/m

≅ 15.8 V/m

6 0
3 years ago
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