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Jlenok [28]
2 years ago
5

Accurate _______ is needed in a valid experiment. Hurry, I need help.

Physics
1 answer:
Lorico [155]2 years ago
4 0

Answer: repeatable

Explanation: Accurate repeatable is needed in a valid experiment. I think

You might be interested in
Will kilogram of hydrogen contains more atoms than kilogram of lead?
Zigmanuir [339]
Yes a kg of hydrogen will have more atoms than a kg of lead, because lead has a higher atomic mass, than hydrogen so it will take more atoms of hydrogen to make a kg than lead
4 0
3 years ago
Two 6 ohm resistors in parallel gives an equivalent resistance of
Mashcka [7]

Answer:

3 ohms

Explanation:

6×6/6+6 =3 .............

3 0
3 years ago
To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
vichka [17]

Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

3 0
3 years ago
An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.7 s. A passenger in the elevator i
avanturin [10]

Answer: Tension = 53.6N

Explanation:

Given that

Height h = 1 m

Time t = 1.7 s.

Mass m = 5.1 kg 

From the equation of the motion we can get the acceleration of the elevator:

h = X0+ V0t + at2/2;

Th elevator starts from rest with a constant upward acceleration. Initial velocity Vo = 0, also Xo = 0; thus

a = 2h/t2 = 2 × 1/1.7^2

a = 0.69 m/s2.

Then we can find the tension in the cord by using the formula

T = mg + ma

= 5.1 (9.8 + 0.69)

= 5.1 × 10.5

= 53.6N

7 0
3 years ago
A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP
worty [1.4K]

Answer:

Explanation:

Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms      

a) Capacity of transformer, Pt = 150KVA = 150,000 W

Input Voltage, Vp = 2.8 KV = 2800 V

Current, Ip = 150000/2800 = 53.57 A

Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms

b) i)   Input current = 53.57 A

ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV

iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW

iv) Power factor = 0.83

8 0
3 years ago
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