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3 years ago

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A defibrillator containing a 18.4 μF capacitor is used to shock the heart of a patient by holding it to the patient's chest. Jus

t prior to discharging, the capacitor has a voltage of 13.2 kV across its plates. How much energy E is released into the patient, assuming no energy losses? E = _____________ J

1 answer:

gregori [183]3 years ago

3 0

Answer:

The energy E is 1603.008 J.

Explanation:

Given that,

Capacitor = 18.4 μF

Voltage = 13.2 kV

We need to calculate the energy

Using formula of energy

$E=W=\int{Q}\ dV$ .....(I)

We know that,

$Q=CV$

Put the value of Q in equation (I)

$E=\int{CV}dV$

On integration

$E=\dfrac{CV^2}{2}$

Put the value into the formula

$E=\dfrac{18.4\times10^{-6}\times(13.2\times10^{3})^2}{2}$

$E=1603.008\ J$

Hence, The energy E is 1603.008 J.

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I am planning to make two dives. The first dive is to 60 feet for 45 minutes, and the second dive is to 60 feet for 35 minutes.

Tpy6a [65]

Answer:

1:04-1:10 hours

Explanation:

You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:

In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
You'll check and keep the letter in that 47 row (S) for future use.
Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.

I hope you find this information useful and interesting! Good luck!

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3 years ago

The fact that some well-known studies have been repeated without finding results consistent with those in the initial report des

ValentinkaMS [17]

Answer:

Replication Crisis

Explanation:

Replication crisis in psychology- It refers to the concerns about credibility of finding the results in psychological science.

6 0

3 years ago

Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

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3 years ago

HELP PLEASE<br><br> Fill in the blank:<br> The ? of water is 1 g/cm3.

ololo11 [35]

I think it’s density

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3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago