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Irina-Kira [14]
3 years ago
10

State 2 defects of primary cells​

Physics
1 answer:
Amanda [17]3 years ago
4 0

local action and polarisation

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Write down the use of hydraulic lift.​
Ksivusya [100]

Answer: The Many Uses for Hydraulic Lifts. Hydraulic lifts are utilized in many extraordinary applications. They may be located in the automotive, shipping, construction, waste removal, mining, and retail industries as they are a powerful approach to elevating and reducing people, goods, and equipment.

8 0
4 years ago
Please help me with this question
valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

4 0
3 years ago
_____ you remember to put the lid back on the jar of mayonnaise​
Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

7 0
3 years ago
Write about Geothermal energy????????????
svlad2 [7]
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8 0
3 years ago
A planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbit
puteri [66]
The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:
\frac{r_1^3}{T_1^2}= \frac{r_2^3}{T_2^2}
where
r_1 is the orbital radius of the first satellite
r_2 is the orbital radius of the second satellite
T_1 is the orbital period of the first satellite
T_2 is the orbital period of the second satellite

If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
T_2 =  \sqrt{T_1^2 ( \frac{r_2}{r_1} )^3} = \sqrt{(18.0 h)^2 ( \frac{3\cdot 10^7 m}{2 \cdot 10^7 m} )^3}  = 33.1 h
8 0
3 years ago
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