What is the potential energy of a 2 kilogram potted plant that is on a 1 meter high plant stand?

The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee

aniked [119]

Answer: 15m/s

Explanation: Average Velocity is vector describing the total displacement of an object and the time taken to change its position. It is represented as:

$v=\frac{\Delta x}{\Delta t}$

At t₁ = 1.0s, displacement x₁ is:

$x(1)=25+3(1)^{2}$

x(1) = 28

At t₂ = 4.0s:

$x(4)=25+3(4)^{2}$

x(4) = 73

Then, average speed is

$v=\frac{73-28}{4-1}$

v = 15

The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s

5 0

2 years ago

A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. $Q$ , the amount of charge stored in this capacitor, will stay the same.

The formula $\displaystyle Q = C\, V$ relates the electric potential across a capacitor to:

$Q$ , the charge stored in the capacitor, and
$C$ , the capacitance of this capacitor.

While $Q$ stays the same, moving the two plates apart could affect the potential $V$ by changing the capacitance $C$ of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

$\displaystyle C = \frac{\epsilon\, A}{d}$ ,

where

$\epsilon$ is the permittivity of the material between the two plates.
$A$ is the area of each of the two plates.
$d$ is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of $\epsilon$ . Neither will that change the area of the two plates.

However, as $d$ (the distance between the two plates) increases, the value of $\displaystyle C = \frac{\epsilon\, A}{d}$ will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula $\displaystyle Q = C\, V$ can be rewritten as:

$V = \displaystyle \frac{Q}{C}$ .

The value of $Q$ (charge stored in this capacitor) stays the same. As the value of $C$ becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.

3 0

3 years ago

Describe your acceleration as you skate down your neighborhood sidewalk.

Paraphin [41]

Answer: when you increase or decrease your speed.

Explanation:

Moving a skate at rest, you need to apply force in order to cause acceleration.

F = ma Where

F = force applied

m = mass of the skate

a = acceleration

The initial velocity u will be equal to zero and the skate will acceleration to a certain velocity.

as you skate down your neighborhood sidewalk, you will accelerate when you increase your speed. Because

Acceleration is the rate of change of velocity. That is,

Acceleration = change in velocity/ time.

And also, you will decelerate when you reduce the speed or velocity down your neighborhood sidewalk.

6 0

3 years ago

60 POINTS!!

musickatia [10]

How can one explain and predict the interactions between objects and within a system of objects? ... through electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction. ... Forces at a distance are explained by fields (gravitational, electric, and magnetic) ...

8 0

3 years ago

tu hermano y tu van al bosque a jugar basquetbol un rato, de regreso apuestan que el que llegue al final hará la cama del otro d

daser333 [38]

Answer:

I don't speak Spanish sorry do you speak english and hola

6 0

3 years ago