If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.

6 0

3 years ago

Assume that a lightning bolt can be modeled as a long, straight line of current. If 16 C of charge passes by a point in 1.50 x 1

Reptile [31]

Answer:

$B = 7.9012*10^{-5}T$

Explanation:

To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.

The current is defined by the load due to time, that is to say

$I= \frac{q}{t}$

Where,

$q= Charge$

$t = time$

So the current can be expressed as:

$I = \frac{16}{1.5*10^{-3}}$

$I = 10666.67A$

Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

$B =\frac{\mu_0}{2\pi}\frac{I}{r}$

Where,

$\mu_0 =$ Permeability Constant

I= Current

r= radius

Replacing the values we have

$B=\frac{4\pi*10^{-7}}{2\pi}(\frac{10666.67}{27})$

$B = 7.9012*10^{-5}T$

7 0

3 years ago

An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball

djverab [1.8K]

Answer:

because of the idea that like charges get repulsion as a force.

Explanation:

because you wrap the ball with foil, the negative charges will leave the foil and go into the ball by induction. This leaves the foil as a positively charged particle since its electrons left it for the ball, making the ball a negatively charged particle. but if you bring the negative charge near the foil, the electrons will transfer from that and go into the foil, making it negatively charged. Now, because the ball and the foil have the same charge, they repel. the foil flies off.

7 0

2 years ago

6. An aircraft is is travelling along a runway at a Velocity of 25m/s in It accelerates at a rate of 4m/s² for a distance of 750

zysi [14]

Answer:

Take-off velocity = v = 81.39[m/s]

Explanation:

We can calculate the takeoff speed easily, using the following kinematic equation.

$v_{f}^{2}=v_{i}^{2} +2*a*x$

where:

a = acceleration = 4[m/s^2]

x = distance = 750[m]

vi = initial velocity = 25 [m/s]

vf = final velocity

$v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]$

7 0

3 years ago