How can you give a baloon an electric charge

9. The current i in a wire of circular cross section can be calculated from the current density, J, using the following equation

tino4ka555 [31]

Answer:

Explanation:

i = ∫J dA

A = πr²

dA = 2πr dr

i = ∫J 2πr dr

= ∫3 x 10⁸ x 2πr³ dr

=3 x 10⁸ x 2π ∫ r³ dr

integrating and taking limit from r = .9 R to R

3 x 10⁸ x 2π / 4 [ 2⁴ - .9⁴ x 2⁴] x (10⁻³)⁴

= 4.71 x 10⁸ x .3439 x 10⁻¹² x 2⁴

= 25.9 x 10⁻⁴ A

8 0

4 years ago

Which of the following four circuit diagrams best represents the experiment described in this problem?

Valentin [98]

We don't see any circuit diagrams.

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0

3 years ago

An object with a mass M = 250 g is on a plane inclined at 30º above the horizontal and is attached by a string to a mass m = 150

malfutka [58]

Answer:

0.495 m/s

Explanation:

$T$ = tension force in the string connecting the two objects

$M$ = Mass of the object on inclined plane = 250 g = 0.250 kg

$m$ = Mass of the hanging object = 150 g = 0.150 kg

$a$ = acceleration of each object

From the force diagram, force equation for the motion of the object on the inclined plane is given as

$T - Mg Sin30 = Ma\\T = Mg Sin30 + Ma$

From the force diagram, force equation for the motion of the hanging object on the inclined plane is given as

$mg - T = ma\\T = mg - ma$

Using the above two equations

$Mg Sin30 + Ma = mg - ma$

$(0.250)(9.8) Sin30 + (0.250) a = (0.150) (9.8) - (0.150)a$

$a = 0.6125 ms^{-2}$

$h$ = height dropped by the hanging object = 10 cm = 0.10 m

$v$ = Speed gained by the object

Speed gained by the object can be given as

$v = sqrt(2ah)\\v = sqrt(2(0.6125)(0.20))\\v = 0.495 ms^{-1}$

5 0

3 years ago

A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

lions [1.4K]

Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

(c) 1235.06N and 252.05 as a ratio

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity