1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
Solve this questions please ​
Rudik [331]

Answer:

10.37×10³j

Explanation:

hope this helps you !!!

4 0
3 years ago
While playing her guitar , karen plucks one string with increasing levels of force. What effect does this have on the sound prod
Lena [83]

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.

This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

I = kA^2

so on increasing amplitude the intensity will increase and hence it will produce loud sound

8 0
3 years ago
Read 2 more answers
a force is applied to an object at rest with a mass of 100 kg. the same force is applied to an object at rest with a mass of 1 k
Dafna1 [17]
The object with the mass ok 1kg will move more quickly because it is lighter than the 100kg object
3 0
3 years ago
Other questions:
  • Two masses exert a force of 1,161 N on each other when they are 20 km apart. How much force will these two masses exert on each
    12·1 answer
  • When the net force of opposite forces is zero, the forces are
    5·2 answers
  • Can an object be in motion and not in motion at the same time? Explain.
    14·1 answer
  • Why do most objects tend to contain nearly equal numbers of positive and negative charges?
    5·2 answers
  • An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a
    5·1 answer
  • Consider a car that travels between points A and B. The car's average speed can be greater than the magnitude of its average vel
    6·1 answer
  • In a large restaurant an average 60% customers ask for water with their meal. A random sample of 10 customers is selected. Find
    11·1 answer
  • PLEASE ANSWER QUICKLY
    9·1 answer
  • four forces act on an object of mass 100kg, they are 12N(N), 70N(S), 33N(E), 10N(w). calculate the reseulting acceleration of th
    5·1 answer
  • sports photographers often use large aperture, long focal length lenses. what limitations do these lenses impose on the photogra
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!