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3 years ago

How does mechanical energy relate to the work an object can do

Physics

1 answer:

kifflom [539]3 years ago

6 0

I am 11 year old and u just needed to login in so I can't answer ur questions

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There is no air in space astronauts in space cannot hear sounds from outside their spacesuits explain this

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3 years ago

The energy consumed by a home during a month is 90 kWh, how many Joules are we talking about? a good explanation please is for t

andreev551 [17]

Answer:

3.24×10⁸ J, or 324 MJ

Explanation:

"kWh" is a kilowatt-hour. It's the energy used by 1 kilowatt of power after one hour.

A kilowatt is a kilojoule per second.

90 kWh

= 90 kW × 1 hr

= 90 kJ/s × 1 hr

= 90 kJ/s × 3600 s

= 324,000 kJ

= 324,000,000 J

The energy is 3.24×10⁸ J, or 324 megajoules.

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3 years ago

Elements in Group 18, at the far right of the periodic table, are called what?

alekssr [168]

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3 years ago

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:

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

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