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Igoryamba
3 years ago
5

The total amount of energy contained in an object

Physics
1 answer:
Elenna [48]3 years ago
7 0

The total amount of energy stored in the particles of an object is called its internal energy. The internal energy of an object is made up of the kinetic energy due to the random motion of the particles and the potential energy due to the interactive forces among the particles.

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Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi
steposvetlana [31]

Answer:

W=K_f-K_i

Explanation:

The work done on a particle by external forces is defined as:

W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,

According to Newton's second law F=ma. Thus:

W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\

Acceleration is defined as the derivative of the speed with respect to time:

W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,

Speed is defined as the derivative of the position with respect to time:

W=m\int\limits^{v_f}_{v_i} v \cdot dv \,

Kinetic energy is defined as K=\frac{mv^2}{2}:

W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i

3 0
3 years ago
A piece of warm concrete is placed in a cold-water tank, and energy flows between the concrete and the water. Which way does the
sammy [17]

Answer:

Heat flows from hot to cold objects. When a hot and a cold body are in thermal contact, they exchange heat energy until they reach thermal equilibrium, with the hot body cooling down and the cold body warming up. This is a natural phenomenon we experience all the time.

Explanation:

3 0
2 years ago
4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p
Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A ,   n 1  =  4 .2 mol

Number of moles mono atomic gas B ,   n 2  =  3.2mol

Initial energy of gas A ,   K A  =  9500  J

Thermal energy given by gas A to gas B ,   Δ K  =  600 J

Gas constant   R  = 8.314  J / molK

Let  K B  be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

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3 years ago
By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343
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Answer:

29.8 %

Explanation:

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Answer:

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