How to find power in physics?

Usimov [2.4K]

Answer:heat brings it up then down

Explanation:

4 0

3 years ago

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

4 0

4 years ago

Potassium ions (K+) move across a 7.0 -mm- thick cell membrane from the inside to the outside. The potential inside the cell is

Reil [10]

Explanation:

Relation between potential energy and charge is as follows.

U = qV

or, $\Delta U = q \times \Delta V$

= $1.6 \times 10^{-19} \times 70 \times 10^{-3}$

= $112 \times 10^{-22}$ J

or, = $1.12 \times 10^{20} J$

Therefore, we can conclude that change in the electrical potential energy $\Delta U$ is $1.12 \times 10^{20} J$ .

7 0

3 years ago

The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current

olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

5 0

4 years ago

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

insens350 [35]

Answer: a) 1875m. (b) 8.66s

Explanation:

6 0

3 years ago