All categories

2 years ago

What is one of the principles of charges in a conductor?

Physics

1 answer:

nevsk [136]2 years ago

3 0

Answer:

A conductor allows free charges to move about within it. The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. Any excess charge will collect along the surface of a conductor. Conductors with sharp corners or points will collect more charge at those points.

Explanation:

You might be interested in

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

4 0

3 years ago

What speed would a fly with a mass of 0.72 g need in order to have the same kinetic energy as a 1250 kg automobile traveling at

Elena L [17]

Kinetic of automobile

Mass m = 1,250 Kg; V = 11 m/s

Formula: K.E = 1/2 mV²

K.E = 1/2(1,250 Kg)(11 m/s)²

K.E = 75,625 J

Speed required for insect to have the same kinetic energy as automobile

Mass of insect = 0.72 g convert to Kg m = 7.2 x 10⁻⁴ Kg

K.E = 1/2 mV² Derive V =?

V = 2 K.E/m

V = √2(75,625 J)/7.2 x 10⁻4 Kg

V = √2.1 x 10⁸ m²/s²

V = 14,491.34 m/s (velocity of insect)

5 0

3 years ago

Si usted trata de contrabandear lingotes de oro llenando su mochila, cuyas dimensiones son de 56 cm * 28 cm * 22 cm, ¿cuál sería

kati45 [8]

Answer:

Mass, m = 665.77 kg

Explanation:

The question says that "If you try to smuggle gold bars by filling your backpack, whose dimensions are 56 cm * 28 cm * 22 cm, what would be its mass? "

The dimension of gold bar is 56 cm×28 cm×22 cm

We know that the density of gold is 19.3 g/cm³

The mas per unit volume of a material is called its density. It can be given by :

$d=\dfrac{m}{V}\\\\m=d\times V\\\\m=(19.3\ g/cm^3)\times (56\times 28\times 22\ cm^3)\\\\m=665772.8\ g\\\\m=665.77\ kg$

So, the mass of the gold bar is 665.77 kg.

5 0

2 years ago

Q: What happens when cold air approaches a body of warm air? Optional Answers: A. The Warm air rises. B. The warm a

artcher [175]

A. The warm air rises! Hope this helps

8 0

3 years ago

Read 2 more answers