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3 years ago

50 points if you help me

Physics

1 answer:

satela [25.4K]3 years ago

8 0

Answer:

X-precipitation

Y(plants)-photosynthesis

Y(animals)-respiration

W-evaporation

Explanation:

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Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

choli [55]

Answer:

$a=16\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

3 0

3 years ago

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

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an e

PIT_PIT [208]

Answer:

$E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force

Explanation:

From the question we are told that

The charge on the small object is $Q = -4.00 \ nC = -4.00 *10^{-9} \ C$

The force is $F = 24 \ nN = 24 *10^{-9} \ N$

Generally the magnitude of the electric field is mathematically represented as

$E = \frac{F}{Q}$

=> $E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}$

=> $E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force

6 0

3 years ago

Which will most likely result in the revision of a theory?

viva [34]

I would say A not 100℅ thou

4 0

3 years ago

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Pls help ASAP now! This is physical science

const2013 [10]

Option A is correct answer

6 0

3 years ago

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