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Veronika [31]
3 years ago
8

A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor th

rough paddles placed on the skin, causing charge to flow through the heart. Assume that the capacitor is originally charged with 5.0 kV .
Part A

What is the charge initially stored on the capacitor?

a. 3×10−9 C
b. 7.5×104 C
c. 7.5×10−2 C
d. 7.5×10−5 C

Part B

What is the energy stored on the capacitor?

a. 1.9×108 J
b. 380 J
c. 190 J
d. 1.9×10−4 J

Part C

If the resistance between the two paddles is 100 Ω when the paddles are placed on the skin of the patient, how much current ideally flows through the patient when the capacitor starts to discharge?

a. 5×105 A
b. 50 A
c. 2×10−2 A
d. 5×10−2 A

Part D

If a defibrillator passes 17 A of current through a person in 90 μs . During this time, how much charge moves through the patient?

If a defibrillator passes 17 {\rm A} of current through a person in 90 {\rm \mu s} . During this time, how much charge moves through the patient?

a. 190 mC
b. 1.5 C
c. 1.5 mC
d. 17 C
Physics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

Part A: Q = 0.075C

Part B: E = 187.5J

Part C: I = 50A

Part D: ΔQ 1.53C

Explanation:

Part A

Q = C×V

Given

C = 15μF = 15×10‐⁶ F, V = 5kV = 5000V

Q = 15×10‐⁶× 5000 = 0.075C

Part B

Energy = E = 1/2 ×CV² = 1/2 × 15×10‐⁶ × 5000² = 187.5J

Part C

Given R = 100Ω

V = IR, I = V/R = 5000/100 = 50A

Part D

I = ΔQ/Δt

Given Δt= 90ms = 90×10-³s, I = 17A

ΔQ = I × Δt = 17×90×10-³ = 1.53 C

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BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w
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Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

initially both are at rest

so here plug in all values

0 = 100 v_{1f} + 2\times 10

v_{1f} = -0.20 m/s

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

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3 years ago
When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now di
AleksandrR [38]

To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.

We know that by Hooke's law

F=kx

Where,

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k= \frac{F}{x}

k= \frac{mg}{x}

k= \frac{(0.35)(9.8)}{12*10^{-2}}

k = 28.58N/m

Perioricity in an elastic body is defined by

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Where,

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4 0
2 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
2 years ago
What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w
shtirl [24]

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

3 0
3 years ago
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