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pantera1 [17]
3 years ago
10

Explain why astronomers long ago believed that space must be filled with some kind of substance (the “aether”) instead of the va

cuum we know it is today.
Physics
1 answer:
professor190 [17]3 years ago
6 0

Answer:Explained

Explanation:

The scientist of that time could not believe that the electromagnetic waves do not require a medium to travel through so they made a space filling substance necessary to travel of electromagnetic waves called aether.

But after the special 0 theory of relativity this idea is discarded and fell out.

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The relationship between mass and inertia is described by newton's second law of motion. true or false
Mariana [72]
False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
4 0
3 years ago
. Write the following in the scientific notation of 0.0000002400m ​
aleksandr82 [10.1K]

Explanation:

<h2><u>Steps </u><u>:</u></h2>
  1. <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
  2. <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
  3. <u>That's</u><u> </u><u>all </u>

<u>hope</u><u> it</u><u> </u><u>help</u>

<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
7 0
2 years ago
A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the
crimeas [40]

Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

7 0
2 years ago
Can you answer these 2 ?
vredina [299]

Answer:

1. 38,500

2. 308,000

Explanation:

This would require a calculator. To find momentum, you multiply mass and velocity. You always want your mass to be measure in kilograms, but that is irrelevant in this question because they already are, it is just something to remember.

4 0
2 years ago
A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr
Molodets [167]

Answer:

Explanation:

Given

P_1=150 kPa

T_1=20^{\circ}C

V_1=0.5 m^3

T_2=140^{\circ}C

P_2=400 kPa

R for Helium R=2.076

c_v=3.115 kJ/kg-K

mass of gas m=\frac{P_1V_1}{RT_1}

m=\frac{150\times 0.5}{2.076\times 293}

m=0.123 kg

Similarly V_2 can be found

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=0.264 m^3

Work done W=\int_{V_1}^{V_2}PdV

W=\frac{P_2V_2-P_1V_1}{n-1}

W=\frac{mR(T_2_T_1)}{n-1}

Since it is a polytropic Process

therefore PV^n=c

P_1V_1^n=P_2V_2^n

(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}

(\frac{0.5}{0.264})^n=\frac{400}{150}

n=\frac{\ln 2.66}{\ln 1.893}

n=1.533

W=\frac{0.123\times 2.076(140-20)}{1.533-1}

W=57.48 kJ    

From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

i.e. Heat is being removed

3 0
3 years ago
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