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3 years ago

Suppose you increase your walking speed from 6 m/s to 14 m/s in a period of 2 s. What is your acceleration? m/s2

Physics

1 answer:

saveliy_v [14]3 years ago

8 0

4.5m/s you count the number between 6 and 14 and get 9 do 9 divided by 2 and you get 4.5 m/s increase. hope this helps.

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A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

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

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0

3 years ago

What is an independent or manipulated variable?

Paha777 [63]

Is the variable you change, independent, I, something I change.

5 0

3 years ago

1 point

Yuliya22 [10]

PE= 3kg x 10N/kg x 10m
= 300J

8 0

3 years ago

During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio

Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

$\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2$

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where v = 0 m/s is the final angular velocity of the wheel when it stops, $\omega_0$ = 95rad/s is the initial angular velocity of the wheel, $\alpha = -29.23 rad/s^2$ is the deceleration of the wheel, and $\Delta \theta$ is the angle swept in rad, which we care looking for: