3 years ago

What is the acceleration of an electron that has moved between -1.5V and 3.0V? Assume is began at rest.

Physics

1 answer:

Virty [35]3 years ago

3 0

Answer:

Insufficient data.

Explanation:

Hi!

The data you have is only potential diference, so you know that the variation in potential energy of the electron when moving from -1.5V to 3.0V, is 4.5 V.

But you cannot know the acceleration. For that you need to know the electric field, so you can calculate force.

You might be interested in

Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp

xz_007 [3.2K]

The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.

6 0

3 years ago

Which two quantities below can be used to calculate average speed?

defon

Change of position and time

6 0

3 years ago

the half-life of iodine-131 is 8.1 days. how much time had passed if i only have one-fourth of the original sample?

morpeh [17]

Answer:

16.2 days

Explanation:

Find the number of halflives:

1/2 * 1/2 = 1/4 so <u>two</u> halflives have passed

2 * 8.1 days = 16.2 days

8 0

1 year ago

Taste and smell would be considered

natka813 [3]

One of your 5 senses because your 5senses are hearing, smelling, seeing, taste, touching

8 0

4 years ago

A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

Aloiza [94]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

$f=\dfrac{R}{2}$

$f=\dfrac{-31}{2}$

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

4 0

3 years ago