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Ivahew [28]
2 years ago
8

a) The student hypothesizes that a greater fraction of kinetic energy is lost from the system during the collision when the spee

d of the first cart is greater Briefly state one reason the hypothesis might be correct
Physics
1 answer:
Mashcka [7]2 years ago
6 0

Answer with Explanation:

One hypothesis above would probably rely on the situation called <em>"inelastic collision." </em>The two carts are travelling towards each other at different speed. So, this means that once they collide, the kinetic energy will not be conserved but will be transferred as <em>sound energy, thermal energy or material deformation</em>. The impact will be great, thus a <u>greater fraction of kinetic energy will be lost from the system.</u>

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A doctor travels to the east from city a to city b 75 km in 1.0h and returns back in another hour
Nadya [2.5K]
The doctor's average speed for the whole trip is 75 km per hour. His average velocity for the whole trip is zero.
4 0
2 years ago
Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with
ella [17]

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

so

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= -  13.6 x2² / 8²

= -  0.85 eV .

3 0
2 years ago
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0
2 years ago
Compressional stress on rock can cause strong and deep earthquakes, usually at _____.
valentinak56 [21]
The answer is reverse faults. 
7 0
2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0
2 years ago
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