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alukav5142 [94]
3 years ago
12

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of t

he heart muscle between the pacemaker’s electrodes is 550Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.
Physics
1 answer:
9966 [12]3 years ago
6 0

Answer: a) 0.11 Amp b) 0.000429 Joules c) 0.000514 watt

Explanation:

a)current

I=V/r I=6.0/550 =1.1*10^-2 A

b) energy

E= 1.1*10^-2 x (6.0) x 0.00065s(<-- remember to convert ms to seconds) = 0.000429J

C) Average power

P = (0.000429 x 72) / 60

P = 0.000514 W

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An object is moving in a straight line at constant speed. A resultant torce begins to act upon
alekssr [168]

Answer: The velocity magnitude or the velocity direction chages.

Explanation:

According to Newton's second law of motion, the acceleration of a system moved in same direction and is also directly proportional to the external force which acts on it while inversely proportional to the mass. The formula is: a = F/m

Based on the question, since the object obtains acceleration, then it can be infered that there will be changes in the velocity magnitude or the direction as a result of the motion.

3 0
3 years ago
A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a distance d. For the next loading, the spring is
Aliun [14]

Answer:

the work required for the loading of second dart is 64 times greater as work required for loading the first dart.

Explanation:

k = spring constant of the spring loaded toy dart gun

x₁ = compression of spring to load the first dart = d

x₂ = compression of spring to load the second dart = 8 d

E₁ = Work required to load the first dart

E₂ = Work required to load the second dart

Work required to load the first dart is given as

E₁ = (0.5) k x₁² = (0.5) k d²

Work required to load the second dart is given as

E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²

E₂ = 64 E₁

So the work required for the loading of second dart is 64 times greater as work required for loading the first dart

8 0
4 years ago
The atomic number of an element is also the number of
lilavasa [31]
Answer: A, protons

Why: it just is
8 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
What is the current in a 160V circuit if the resistance is 20Ω
Mkey [24]

Answer:

I =V/R

I= 160/20 = 8 Amperes

3 0
3 years ago
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