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Kamila [148]
3 years ago
6

The strongman lifts the pig by pulling down at position 1. How will the distance that he pulls down compare to the distance that

the pig moves up?
Question options:


The distance that the pig moves up will be more than the distance that the strongman pulls down.


The distance that the pig moves up will be the same as the distance that the strongman pulls down.


The distance that the pig moves up will be less than the distance that the strongman pulls down.


It is impossible to tell from this picture.

Physics
2 answers:
Elena L [17]3 years ago
7 0

Answer:

i believe its the 3rd option

Explanation:

if he pulls it down the pig will be level or higher than he pull down

Gennadij [26K]3 years ago
5 0

Answer:

C

Explanation:

The distance that the pig moves up will be less than the distance that the strongman pulls down.

Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.

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Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
Which of the following statements is false?
vladimir1956 [14]

An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.

Option B

<u>Explanation</u>:

In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.

Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.

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3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

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Force bc it says the ability to make stuff happen
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