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3 years ago

What is the average power output (in W) of a heart defibrillator that dissipates 435 J of energy in 10.5 ms?

Physics

1 answer:

Yuki888 [10]3 years ago

7 0

Answer:

The power is $P = 41429 \ W$

Explanation:

From the question we are told that

The energy is $E = 435 \ J$

The time taken is $t = 10.5 \ ms = 10.5 *10^{-3} \ s$

Generally the the average power is mathematically represented as

$P = \frac{E}{t}$

=> $P = \frac{ 435}{ 10.5*10^{-3}}$

=> $P = 41429 \ W$

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A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh

MA_775_DIABLO [31]

Answer:

The resistance is $3.5\times10^{-4}\ \Omega$

Explanation:

Given that,

Diameter of tube = 8.5 mm

Length = 8 cm

Resistivity = 2.5 m

We need to calculate the resistance

The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

$R = \dfrac{\rho\times l}{A}$

Where, $\rho$ =resistivity

l = length

A = area of cross section

Put the value into the formula

$R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}$

$R=3526.32\ \Omega$

$R=3.5\times10^{-4}\ \Omega$

Hence, The resistance is $3.5\times10^{-4}\ \Omega$

6 0

3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$