To determine an epicentral distance scientists consider the arrival times of what wave types

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

7 0

3 years ago

A project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow fo

Ksju [112]

Answer:

A. The project's energy costs will decrease

Explanation:

Since the project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow for on-site renewable energy.

Hence, the project's energy costs will decrease very well because it's implementing both of these strategies;

- Area with demand-response program.

- On-site renewable energy.

5 0

3 years ago

United States citizens have many ways to participate in national life. The most common form of participation is?

Travka [436]

Answer:

You would have to give better explanation on subject.

Explanation:

3 0

2 years ago

What is the difference between charging by contact and charging by induction in terms of electron transfer.

Veronika [31]

Answer:

the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch

Explanation:

There are three methods of charging an object:

- Charging by friction: this is done by rubbing an object against another object. An example is when a plastic rod is rubbed with a wool cloth. When this is done, electrons are transferred from the wool to the rod, so both objects remain charged at the end of the process

- Charging by contact: this is done by putting in contact a charged object with a neutral, conducting object. In this case, the charges are transferred from the charged object to the neutral object; at the end of the process, the neutral object will also have a net electric charge, so it will be also charged.

- Charging by induction: in this case, we take a charged object, and a neutral object, and we place the two objects close to each other, but without touching. Let's assume that the charged object is negatively charged: in this case, the positive charges in the neutral object are attracted towards the negative charges of the charged object, while the negative charges of the neutral object are repelled away. As a result, the positive and negative charges in the neutral object split apart. If the object is connected to the ground, then negative charges move away, so the neutral object will remain positively charged.

Therefore, the main difference between charging by contact and charging by induction is that in the first case, the two objects are touching, while in the second case, the two objects do not touch.

5 0

3 years ago

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

$Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta$

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

$E = (3)(230\ m/s)(0.61\ T)Sin90^o$

3 0

3 years ago