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Scilla [17]
3 years ago
8

The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc

ting gel is used in conjunction with a defibrillator to overcome skin resistance, so that the main resistance of the body to an applied voltage is solely due to the internal resistance of the body, which is roughly around 300 ohms, what voltage is required for defibrillation? (Note, the actual value used is 10-100 times this since the skin's large resistance cannot necessarily be neglected.)
Physics
1 answer:
algol [13]3 years ago
5 0

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

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Helppppppppppppppppppppppppp
andrew11 [14]

Answer:

value of x=35°

y=72.5°

hope it helps you

make me brainliest plz

6 0
3 years ago
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
3 years ago
A 0.290 kg block on a vertical spring with a spring constant of 5.00 ✕ 103 N/m is pushed downward, compressing the spring 0.110
True [87]

Answer:

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant : K= 5 x 10 to the 3rd power n/m

compression x = 0.10 m

Mass of block m= 0.250 kg

Here spring potential energy converted into potential energy,

mgh = 1/2 kx to the 2 power

For finding at what height it rise,

0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power

h= 10.20

Therefore, the height at point of release is 10.20 m

4 0
3 years ago
In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first
Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

3 0
3 years ago
Read 2 more answers
A jet travels at 489 meters per second how long does it take to travel 235 m? answer in seconds
Sladkaya [172]

if in 1 second the jet goes 428 meters and you want to know how long it takes to go 237 meters you just divide 237/428  

it can be written like this  

_1_ = __x__  

428 237  

cross multiply  

428 times x = 1 times 237  

428x=237  

divide by 428 on both sides  

x= 237/428  

x=0.5537 seconds

7 0
4 years ago
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