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1 year ago

7

what will be the resistivity of a metal wire of 2m length and 0.6mm in a diameter ,if the resistance of the wire is 50ohm . find

the resistance of the same wire of 2m length and 0.3mm thickness

1 answer:

Maru [420]1 year ago

5 0

Answer:Resistivity of the wire is 7.065 × 10^-8 Ω m.

Explanation:

I have no idea i just googled it

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Write difference between upstroke and downstroke of water pump

lakkis [162]

Explanation:

Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.

7 0

2 years ago

Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont

sergeinik [125]

<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable

The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air

Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body

∴ For a lumped system analysis Biot number should be less than 0.1

4 0

2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

1 year ago

Two teams are pulling a heavy chest, located at point X. The teams are 4.6 meters away from each other. Team A is 2.4 meters awa

Sladkaya [172]

For a better understanding of the question, please see attached picture that I've created.
We need to identify the other two angles by applying the Law of sine where we are given with the following values:
a = 3.2 unit
b = 2.4 unit
c = 4.6 unit
∠A = unknown
∠B = unknown
∠C = 110°

Solving for ∠B, we have:
sin C / c = sin B / b
sin 110 / 4.6 = sin B / 2.4
sin B = 2.4*sin 110 / 4.6
B = 29.36°

Solving for ∠A, we have:
sinC/c = sinA/ a
sin110 / 4.6 = sinA / 3.2
sinA = 3.2*sin110/ 4.6
A = 40.82°

Therefore, the missing angles are ∠A=40.82° and ∠B=29.36°.

3 0

2 years ago

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Two objects, labeled A and B, are the same size. Object A has a density of 1.21 g/cm3. Object B has a density of 1.37 g/cm3. Bot

Sedaia [141]

Neither object will float in water.

They both have densities greater than 1.0 g/cm³,
so both will sink to the bottom of the beaker.

Like rocks.

4 0

2 years ago

Read 2 more answers