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dezoksy [38]
3 years ago
11

Increase the accurancy of a scientific experiment

Physics
2 answers:
ArbitrLikvidat [17]3 years ago
7 0
The best way to increase the accuracy of a scientific experiment is to have a detailed plan of action for conducting the experiment. You should also repeat the experiment several times to ensure that you get the same result(s) consistently.<span />
Charra [1.4K]3 years ago
3 0
Have an idea them make it detailed then do it having correct measurements
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A heat pump receives heat from a lake that has an average winter time temperature of 6o C and supplies heat into a house having
Natalija [7]

Answer:

Explanation:

60000 kJ / h = 60000 x 1000 / (60 x 60 )

= 16667 J /s

a)

\frac{Q_H}{W} =\frac{T_H}{T_H-T_C}

where Q_H and W are heat supplied and heat given from  outside source .

Here Q_H = 16667 J/s

\frac{T_H}{T_H-T_C}=\frac{298}{298-279}

= 15.684

Q_H = 16667 J/s

16667 / W = 15.684

W= minimum power supplied  = 1062.7 W.

b )

If  T_C=283K

\frac{T_H}{T_H-T_C}=\frac{298}{298-283}

=19.87

\frac{Q_H}{W} = 19.87

W=\frac{Q_H}{19.87}

W=\frac{16667}{19.87}

= 838.8 W .

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3 years ago
Provide an example of a material that transmits, reflects, and absorbs light simultaneously?
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A Mirror



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3 years ago
When pressing the accelerator on a car, the ___________ energy of the fuel is transformed into the _____________ energy that mak
lesya [120]
B. Chemical; Mechanical
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3 years ago
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In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x
Ostrovityanka [42]

Answer:

(a) increase

Explanation:

On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.

The change in the potential energy of a charge field-system can be given as:

\delta U= -q(EdsCos \theta)

where;

q = positive test charge

E = Electric field

ds = displacement between thee charge positions

θ  = Angle between the electric field and the displacement.

Given that:

Charge of the particle = -q

displacement = (60.0 -20.0)cm = 40.0 cm

θ = 0

Replacing our values in the above equation, we have:

\delta U = -(-q)(60Cos 0)

\delta U = qEds

Since the potential energy of the system is positive, therefor the electric potential energy also increases.

5 0
4 years ago
Could somebody double check my work???
irina1246 [14]

Answer:

I think it's correct. Hope it helps.

5 0
3 years ago
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