What is the minimum total energy released when

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

7 0

Answer:

(1) $1.64 \times 10^{-13} J$

Explanation:

Energy released in this process is given by

$Q = \Delta m c^2$

here we have

$\Delta m = 9.1 \times 10^{-31} kg$

since here mass of two electrons converted into energy so we have

$Q = 2(9.1 /times 10^{-31})(3\times 10^8)^2$

$Q = 1.64 \times 10^{-13} J$

so here energy released is the energy of rest mass energy due to two charges i.e. electrons and positrons

Ganezh [65]3 years ago

6 0

What is the minimum total energy released when an electron and its antiparticle (positron) <span>annihilate each other?
The </span>minimum total energy released when an electron and its antiparticle (positron) annihilate each other is <span>2.73 × 10^–22 J. The answer is number 4.</span>

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A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .