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3 years ago

Is directly proportional to the length and resistivity of the conductor.

Physics

2 answers:

aliya0001 [1]3 years ago

8 0

Answer:

B) resistance

Explanation:

the resistance of a wire is proportional to its length, and inversely proportional to its cross-sectional area.

Andrews [41]3 years ago

4 0

The resistance of conductor is directly proportional to length and resistivity.

Answer: Option B

Explanation:

Resistance is present in an electrical object or rather introduced to counter check the flow of the current. It is similar to the mechanical friction in the conceptualization of its existence and effect.

It is directly proportional to the length of the electric circuit. It is being measured and noted in ohm. The formula for resistance is –

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

Where,

R - the Resistance of the wire,

$\rho$ - the resistivity of the wire

L - The length of the wire

A - the area of cross section of the wire

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A child of mass M is swinging on a swing set. The ropes attaching the swing to the top bar have length L. Find the gravitational

myrzilka [38]

Answer:

(a) 0

(b) 10ML

(d) $10ML(1 + sin(\phi))$

Explanation:

(a) When hanging straight down. The child is at the lowest position. His potential energy with respect to this point would also be 0.

(b) Since the rope has length L m. When the rope is horizontal, he is at L (m) high with respect to the lowest swinging position. His potential energy with respect to this point should be

$E_h = mgh = 10ML$

where g = 10m/s2 is the gravitational acceleration.

(c) At angle $\theta$ from the vertical. Vertically speaking, the child should be at a distance of $Lcos(\theta)$ to the swinging point, and a vertical distance of $L - Lcos(\theta)$ to the lowest position. His potential energy to this point would be:

$E_{\theta} = mgh = 10M(L - Lcos(\theta)) = 10ML(1 - cos(\theta))$

(d) at angle $\phi$ from the horizontal. Suppose he is higher than the horizontal line. This would mean he's at a vertical distance of $Lsin(\phi)$ from the swinging point and higher than it. Therefore his vertical distance to the lowest point is $L + Lsin(\phi) = L(1 + sin(\phi))$

His potential energy to his point would be:

$E_{\phi} = mgh = 10ML(1 + sin(\phi))$

5 0

3 years ago

A bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg. What is the

kiruha [24]

Answer:

m1 = 20g (= 0.02 kg)

Mass of pistol, m2 = 2 kg

Initial velocity of the bullet (u1) and pistol (u2) = 0

Final velocity of the bullet, v1 = +150m s-1

Let v be the recoil velocity of the pistol.

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)

= (3 + 2v) kg m s-1

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s

7 0

3 years ago

It is cold and dry outside. You go down the slide and experience a small electric shock. What charge must you and slide be in or

ruslelena [56]

Answer:

Explanation:

the charge wants to ultimately balance out so either you or the slide gave the other electrons to balance the charges

3 0

3 years ago

Read 2 more answers

Which quantity can be calculated using the equation E=mc2

EastWind [94]

Well, the equation describes the relationship among
three physical quantities ...

E . . . energy
m . . . mass
c . . . the speed of light.

Speaking strictly algebraically, if you know any two of them,
you can use the equation to calculate the third one.

5 0

3 years ago

According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

e-lub [12.9K]

Answers:

a) $\rho_{cylinder}= 0.55 g/cm^{3}$

b) $\rho_{liq}= 1.48 g/cm^{3}$

c) When we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

Explanation:

a) According to Archimedes’ Principle:

A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium. This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

Hence:

$W_(cylinder)=B$ (1)

Where:

$W_(cylinder)=m.g$ is the weight of the wooden cylinder, where $m$ is its mass and $g$ gravity.

$B$ is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

We can rewrite (1) as follows:

$m_{cylinder}g=m_{water}g$ (2)

On the other hand, we know density $\rho$ establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:

$\rho=\frac{m}{V}$ (3)

isolating the mass:

$m=\rho V$ (4)

Now we can express (2) in terms of the density and the volume of cylinder and water:

$\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g$ (5)

In this case $V_{water}$ is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).

At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:

$V_{cylinder}=\pi r^{2} h$ (6)

Where $r$ is the radius and $h=30 cm$ the total height of the cylinder.

$V_{water}=\pi r^{2} (h-h_{top})$ (7)

Where $h_{top}=13.5 cm$ is the height of the top of the cylinder above the surface of water and $(h-h_{top})$ is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

$\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g$ (8)

Clearing $\rho_{cylinder}$ :

$\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}$ (9)

Simplifying;

$\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}$ (10)

Knowing $\rho_{water}=1g/cm^{3}$ :

$\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})$ (11)

$\rho_{cylinder}= 0.55 g/cm^{3}$ (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ( $\rho_{cylinder}= 0.55 g/cm^{3}$ ), but the density of the liquid $\rho_{liq}$ is unknown.

Applying again the Archimedes principle:

$\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g$ (13)

Isolating $\rho_{liq}$ :

$\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}$ (14)

Where:

$V_{cylinder}=\pi r^{2} h$

$V_{liq}=\pi r^{2} (h-h_{top})$

Then:

$\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}$ (15)

$\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}$ (16)

$\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}$ (17)

$\rho_{liq}= 1.48 g/cm^{3}$ (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor $\pi r^{2}$ is removed during calculations.

6 0

4 years ago