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Sladkaya [172]
2 years ago
7

I Explain why you feel cold when tap or well water in winter. state with reason.​

Physics
1 answer:
attashe74 [19]2 years ago
3 0

Answer:

We feel cold when tap or well water in winter because heat flows from hot body to cold body.

Explanation:

Our <em>body</em><em> </em><em>is</em><em> </em><em>in</em><em> </em><em>optimal</em><em> </em><em>status</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>hot</em><em> </em><em>body</em><em> </em><em>and</em><em> </em><em>tap</em><em> </em><em>or</em><em> </em><em>we</em><em>ll</em><em> </em><em>water</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>cold</em><em> </em><em>body</em><em>.</em><em> </em><em>Theref</em><em>ore</em><em> </em><em>we</em><em> </em><em>feel</em><em> </em><em>cold</em><em>.</em>

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50/0.04= answer

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
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Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

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6 0
3 years ago
What does the voltage-current graph above show about the relationship between voltage and current, and how do these properties a
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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

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