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kolezko [41]
3 years ago
6

In an investigation that uses the scientific method, which step immediately follows asking a question?

Physics
2 answers:
Vesnalui [34]3 years ago
6 0

Answer:

The correct answer is option B , Making a hypothesis

Explanation:

In general, all scientific researches  follow a sequential pattern, which is as follows-

a) Making observations

b) Asking a question  

c) creating a hypothesis

d) Designing an experiment

e) Testing the hypothesis

f) presenting the results

The hypothesis is basically based on the question asked and consists of two scenarios either its acceptance or its rejection. After the hypothesis is frame the researcher design the methodology to test the hypothesis. Once the hypothesis is tested, obtained results are presented.  

almond37 [142]3 years ago
4 0
Creating a hypothesis
We attempt to explain the question raised and then design an experiment according to the question and test the variables.
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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
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Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

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Mass of the cart = 146 kg

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3 years ago
What is required for an electric charge to flow through a wire?
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3 years ago
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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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Answer:

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