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2 years ago

Explain how horizontal motion can be uniform while vertical motion is accelerated.

1 answer:

7 0

In projectile motion horizontal motion has zero acceleration and when there is no air resistance acting on the object the motion will be uniform. Motion in the vertical direction is accelerated downwards to to the acceleration due to gravity

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

The Kelvin scale is the most common temperature scale used in what

dalvyx [7]

It's mostly used in CHEMICAL PROCESSES.

7 0

3 years ago

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Which statement is TRUE concerning the particle arrangement of a solid

Anni [7]

It's packed together

8 0

3 years ago

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The net electric charge of an amber rod which has been rubbed with fur is called negative Group of answer choices because amber

bonufazy [111]

Answer:

The right option is option E. None of the answer choices given are totally correct.

Explanation:

All insulators normally have an equal amount of positive and negative charges distributed on their surface.

The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.

The fur in turn becomes positive because it has more positive charges than negative on its surface.

So, the convention allows the now rubbed off amber rod to be called negative.

So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!

Hope this helps!!

3 0

3 years ago

In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu

laiz [17]

Answer:

a) Eₓ = - A y + 2B x , b) Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

E = - dV / ds

a) Let's find the electric field on the x axis

Eₓ = - dV / dx

dV / dx = A y - B 2x

Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

Ey = - dV / dy

dV / dy = A x + C

Ey = -Ax –C

c) the electric field on the z axis

dv / dz = 0

Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

0 = = - A y + 2B x

y = 2B / A x

Axis y

0 = -Ax –C

.x = -C / A

We substitute this value in the previous equation

.y = 2B / A (-C / A)

.y = 2 B C / A2

The correct answer is 3

6 0

4 years ago