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HACTEHA [7]
2 years ago
12

Is grade 10 harder or 9? I passed 9 grade thise year by 89,36 my mark..im scared of high school plis help and some advice ?

Physics
1 answer:
Liula [17]2 years ago
4 0
Listen in all of your classes and write down notes on EVERYTHING even if it may not seem important it will be useful in the future! :)
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Which parts of the electric circuit considered as fuse ​
Marizza181 [45]

Answer:

Rewirable or Kit – Kat Type Fuses are a type of Low Voltage (LV) Fuses. They are most commonly used in house wiring, small industries and other small current applications. Rewirable Fuses consists of two main parts: a Fuse Base, which contains the in and out terminal, and a Fuse Carrier, which holds the Fuse Element.

3 0
2 years ago
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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

4 0
3 years ago
Which of the following is true about wedges?
timofeeve [1]

Answer:

wedges are a type of inclined plane.

Explanation:

i just answered :) :) :)

8 0
3 years ago
HELP!!
tresset_1 [31]

Answer:

So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3

3 0
2 years ago
A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine
Goryan [66]

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

Heat energy production = 20.0 KJ/s

Coefficient of performance,  K = 7

also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

P(in) is the input power

Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

4 0
2 years ago
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