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3 years ago

The radius of a conducting wire is doubled .What will be the ratio of its new specific resistance to the old one?

1 answer:

7 0

The equation for the resistance R is: R=ρ*(l/A), where, ρ is electrical resistivity, l is the length of the conductor, and A is the surface area.

The initial surface area is:

A=r²π, then when we double the radius we get:

A₁=(2*r)²π=4*r²π=4*A

Initial resistance is: R=ρ*(l/A).

When we double the radius, resistance is: R₁=ρ*{ l / (4*A) }

The ratio of the new resistance to the old one:

R₁/R=[ρ*(l/A)] / [ ρ* { l / (4*A) } ] = ρ, l and A cancel out and we get:

R₁/R=(1/1)/(1/4)=4/1

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When scientists state that energy transformation are inefficient, do they mean energy is destroyed? Explain why.

TiliK225 [7]

no. energy is not created or destroyed, just transformed.

a hot light source is less efficient than a cold one ... heat is not light ...

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3 years ago

When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final p

Elina [12.6K]

As per question the initial states of the gases are given as

INITIAL STATE: FINAL STATE:

$p_{1} =4.90106 pa$ $p_{2} =1.06106 pa$

$v_{1} =v[say]$ $v_{2} =3v[say]$

$T_{1} =20 degree celcius =293 K$ $T_{2} =?$

AS per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

$\frac{p_{1} v_{1} }{T_{1} } =\frac{p_{2}v_{2} }{T_{2} }$

Hence $T_{2} =\frac{p_{2} v_{2}T_{1} }{p_{1} v_{1} }$

= $\frac{1.06106*3v*293}{4.90106*v}$

=190.3 K [ANS]

6 0

2 years ago

How many years are in a light year?

Semenov [28]

How long would it take the space shuttle to go one light-year? The shuttle orbits the Earth at about 5 miles per second (18,000 mph). Light travels at 186,000 miles per second, which is about 37,200 times faster than the shuttle. So the shuttle would need about 37,200 years to go one light-year.

8 0

2 years ago

Read 2 more answers

Ted throws an object straight up into the air with an initial velocity of 54 ft/s from a platform that is 40 ft above the ground

elena-s [515]

Answer:

The time it will take for the object to hit the ground will be 4.

Explanation:

You have:

h(t)=−16t²+v0*t+h0

Being v0 the initial velocity (54 ft/s) and h0 the initial height (40 ft) and replacing you get:

h(t)=−16t²+54*t+40

To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:

0=−16t²+54*t+40

Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:

$\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}$

In this case you have that:

a=-16
b= 54
c= 40

Replacing in the expression of the calculation of roots you get:

$\frac{-54+\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}$ Expresion (A)

and

$\frac{-54-\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}$ Expresion (B)

Solving the Expresion (A):

$\frac{-54+\sqrt{5476 } }{2*(-16)}= \frac{-54+74}{2*(-16)}=\frac{20}{2*(-16)}=\frac{20}{-32}= -\frac{5}{8}$

Solving the Expresion (B):

$\frac{-54-\sqrt{5476 } }{2*(-16)}= \frac{-54-74}{2*(-16)}=\frac{-128}{2*(-16)}=\frac{-128}{-32}= 4$

These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then <u><em>the time it will take for the object to hit the ground will be 4.</em></u>

6 0

2 years ago

Your physical education teacher has set up a modified shuttle run event for a class competition. From the starting position, you

motikmotik

+14-17+16-28 = -17+16-14 = -1-14=-15 ... west, and out of breath

3 0

2 years ago