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2 years ago

If the potential difference across the bulb in a camping lantern is 9.0 V, what is the

Physics

1 answer:

Gre4nikov [31]2 years ago

4 0

Answer:

Explanation:

The potential difference across the terminal as the same and thats because we are assuming that the source has no internal resistance.

Internal resistance are usually little resistances in the supply.

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HELP ME PLEASE WITH MY HOMEWORK ILL GIVE BRAINLIEST

nadezda [96]

Answer: Mass

Explanation:

8 0

2 years ago

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

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6 0

3 years ago

Give reason : the action and reaction do not lead to equilibrium ?

Natalija [7]

the action and reaction do not lead equilibrium if action and reaction force react on different objects

4 0

2 years ago

An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports

finlep [7]

Option D is correct. An arch carries the thrust of weight to its sides with a post-and-lintel.

An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it

In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.

Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its sides with a post-and-lintel.

To learn more about the arch refer to the link;

brainly.com/question/18162421

6 0

2 years ago

A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t

sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, $\rho=2.44\times 10^{-8}\ \Omega-m$

Resistance in terms of resistivity is given by :

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

Also, V = IR

So,

$\dfrac{V}{I}=\dfrac{\rho l}{A}$

A is area of wire,

$\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}$ , r is radius, r = d/2 (diameter=d)

$\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m$

Out of four option, near option is (C) 17 μm.

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2 years ago