What is the hottest planet in the milky way galaxy

Pachacha [2.7K]

Answer:

Less than the distance

3 0

3 years ago

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

6 0

3 years ago

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int

Inessa05 [86]

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a → a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m → a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

3 0

2 years ago

Which diagram best shows the field lines around two bar magnets attracting each other?

Marizza181 [45]

I think the answer is C

5 0

2 years ago

Which condition is required for Coulomb's law to hold true?

AleksAgata [21]

The correct answer is:
Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
 $F= \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$
where
 $\epsilon_0$ is the permittivity of free space
q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
 $F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$
where $\epsilon_r$ is the relative permittivity, which takes into account the dielectric effects of the material.

7 0

2 years ago