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Ulleksa [173]
3 years ago
13

Factor that changes because of the manipulated variable

Physics
1 answer:
jeyben [28]3 years ago
6 0

Answer:

factor that changes because of the manipulated variable: A. Manipulated variable

xXxAnimexXx

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Now review the difference between the temperatures on Earth and Mars. Also look at their distances from the sun.
aksik [14]

Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away

Explanation: HOPE THIS HELPSS!! ;))

7 0
2 years ago
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In physics what does 7.56 × 5.746 equal ?
lapo4ka [179]

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

7 0
3 years ago
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part t
UNO [17]

Answer:

\alpha=-3.01dB

Explanation:

From the question we are told that:

Sound level intensity

 \triangle I=40dB-80dB

Generally the equation for  intensity level  is mathematically given by

 \alpha=10log_{10}(I/I_x)dB

Where

 I= Intensity measured

 I_x=Threshold\ of\ audibility

 I_x= 10-12 W / m2

 \alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}

 \alpha= 10 log10 \frac{I_1}{I_2}

 \alpha=10 log10\frac{40}{80}

 \alpha=-3.01dB

7 0
3 years ago
How many 1140 nm long molecules would you have to line up end to end to stretch a distance of 158 miles?
dezoksy [38]

Answer:

221754385964.9123

Explanation:

Convert miles to nanometer

1 mile = 1.6 km

1 km = 1×10³×10³×10³×10³ nm

1 mile = 1.6×10¹² nm

So,

158 miles = 158×1.6×10¹² = 252.8×10¹² nm

Length of each molecule = 1140 nm

Number of molecules = Total length / Length of each molecule

\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123

There are 221754385964.9123 number of molecules in a stretch of 158 miles

3 0
3 years ago
Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field
Nuetrik [128]

Answer:

q=6.22*10^-10C

Explanation:

Two large metal plates of area 0.88 m2 face each other, 4.8 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 80 N/C. Find |q|

E=α/∈, electric field within the plate

α=q/A

A=area of the plate

∈=is the permittivity

substituting , we have

The field magnitude E between them (neglect fringing)

E=q/A∈

q=EA∈

q=0.88*80*8.84*10^-12

q=6.22*10^-10C

3 0
3 years ago
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