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3 years ago

A wire of length L and cross-sectional area A has resistance R.

Physics

1 answer:

Leona [35]3 years ago

6 0

Answer:

The resistance R stretched of the wire if it is stretched to twice its original length is [Δ(2L) / A ]* 2

Explanation:

Original Resistance R = ΔL / A

Given length is doubled

New length L' = 2L

New Volume is constant, resistivity and density of material is also assumed to be constant

AL = A'L'

AL = A'(2L)

A' = AL / 2L

A' = A / 2

∴ new Resistance R' = ΔL' / A'

= Δ(2L) / A/2

= [Δ(2L) / A ]* 2

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If the focal length of a concave mirror is 18cm, find its radius of curvature.

Galina-37 [17]

Given :

The focal length of a concave mirror is 18 cm.

To Find :

The radius of curvature of the concave mirror.

Solution :

We know,

$\text{Focal length}=\dfrac{\text{Radius of curvature}}{2}\\\\F=\dfrac{R}{2}\\\\R = 18\times 2\ cm\\\\R = 36 \ cm$

Therefore, the radius of curvature of concave mirror is 36 cm.

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3 years ago

A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance

Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, $m_b=11\ g=0.011\ kg$

Mass of the pendulum, $m_p=19\ kg$

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We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

$\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)$

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

$m_bv=(m_p+m_b)V$

Put the value of V from equation (1) in above equation as :

$v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s$

So, the bullet's initial speed is 243.21 m/s.

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3 years ago

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3 years ago

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balandron [24]

A or possibly C because the other options have nothing to do with the size of the vibration. If i was you I would answer with A

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3 years ago

Increasing the number of loops in a electromagnet or solenoid will cause it to be stronger

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True

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2 years ago