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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 spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two

vampirchik [111]

Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

R₁ = 3.59 cm

R₂ = 7.22 cm

m₁ = m₂ = m

K₁ = K₂

We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴ K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² (I)

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴ K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂² (II)

∵ K₁ = K₂

⇒ 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒ ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒ (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒ (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒ (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒ (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒ (ω₁ / ω₂) = 1.9079

8 0

3 years ago

A 5-mm-thick stainless steel strip (k = 21 W/m·K, rho = 8000 kg/m3, and cp = 570 J/kg·K) is being heat treated as it moves throu

Snezhnost [94]

Answer:

1170 Kelvin/meter

Explanation:

See attached pictures.

8 0

3 years ago

Read 2 more answers

How can a gas become a good conductor? Simpler answers would be helpful.

Alex_Xolod [135]

A gas has to become ionisied in order to become a conductor. It must have a chain reaction in which atoms in it became unstable, in which they loose stabile electronic configuration. In order for a gas to become a conductor, it must have free particles, and it can happen only in ionisied gas.

7 0

3 years ago

Gravity prevents planets from flying off at a

Bad White [126]

Answer:elliptical orbit

Explanation:

7 0

3 years ago

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$

Replacing we get that I is:

$I=(200N*0,33m)/(0,936rad/s^2)$

Then $I=70,513kgm^2$

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:

$I=(1/2)*(m*r^2)$

4 0

3 years ago