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Lostsunrise [7]

3 years ago

11

A wire 40cm long and of diameter 0.60mm has a resistance of 1.5 ohm what is the resistivity of the material of which it is made

taking pie to be 3.14

1 answer:

zavuch27 [327]3 years ago

3 0

Answer:

The resistivity of the wire is $1.05\times 10^{-6}\ \Omega-m$

Explanation:

We have,

Length of a wire is 40 cm or 0.4 m

Diameter of a wire is 0.60 mm

Radius is 0.3 mm or 0.0003 m

Resistance of a wire is 1.5 ohm

Now we need to find the resistivity of the material of which it is made. The resistance of a wire in terms of its resistance, length and area is given by :

$R=\rho \dfrac{l}{A}$

$\rho$ = resistivity

$\rho=R \dfrac{\pi r^2}{l}\\\\\rho=1.5\times \dfrac{3.14\times (0.0003 )^2}{0.4}\\\\\rho=1.05\times 10^{-6}\ \Omega-m$

So, the resistivity of the wire is $1.05\times 10^{-6}\ \Omega-m$ .

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The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m (2,716.5 feet) and has more than 160 stories.

Elena L [17]

Answer:

h = 599.5 m

Explanation:

Given,

height of structure = 828 m

weight of the tourist = 184 lb

= 184 x 0.45359 = 83.43 Kg

Potential energy = 187000 J

PE = m gd

$d = \frac{PE}{mg}$

$d = \frac{187000}{83.43\times 9.81}$

h = 228.5 m

Height of the room above the ground.

h = 828 - 228.5

h = 599.5 m

Height of the floor above ground is equal to 599.5 m.

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What are 2 ways organisms would be affected by water pollution?

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Not having clean water to drink or bath. Also some organizations that live in the water might die because the water is polluted and that would be toxic form them

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

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When the heat source is removed from a fluid, convection currents in the fluid will eventually ____________then stop.

Blizzard [7]

When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid. When all of the fluid is at the same temperature, convection currents will stop.

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3 years ago

A cyclist turns a corner with a radius of 50m at a speed of 10m/s. What is the cyclist's acceleration?

miv72 [106K]

2m/s²

Explanation:

Given parameters:

Radius = 50m

Speed = 10m/s

Unknown:

Acceleration of the cyclist

Solution:

The acceleration of the cyclist is directed inside of the corner because his motion is inward. This is a form of centripetal acceleration;

Centripetal acceleration is given by;

a = $\frac{v^{2} }{r}$

v is the velocity

r is the radius

a = $\frac{10^{2} }{50}$ = 2m/s²

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

3 0

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