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Debora [2.8K]
3 years ago
5

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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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

3 0
3 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®
Furkat [3]

C. Kevlar

:) ++ * ++ !

5 0
3 years ago
Read 2 more answers
PLEASE PLEASE HELP!!!!!!
Whitepunk [10]

Answer:

the answer is B

Explanation:

The atomic mass of an atom is the sum of the protons plus neutrons it has.

7 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
How can we tell when forces are acting on an object (science)
iren [92.7K]

If an object's speed changes, or if it changes the direction it's moving in,
then there must be forces acting on it. There is no other way for any of
these things to happen.

Once in a while, there may be <em><u>a group</u></em> of forces (two or more) acting on
an object, and the group of forces may turn out to be "balanced".  When
that happens, the object's speed will remain constant, and ... if the speed
is not zero ... it will continue moving in a straight line.  In that case, it's not
possible to tell by looking at it whether there are any forces acting on it. 


4 0
3 years ago
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