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enyata [817]
3 years ago
7

Physical isolation of atrial muscle cells from ventricular muscle cells is a function of which of the following?

Physics
1 answer:
solong [7]3 years ago
5 0

Answer:

The cardiac skeleton

Explanation:

The cardiac skeleton is not a "true" skeleton, but it provides structure and support for the heart, as well as isolating the atrial muscle cells from ventricular muscle cells.

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When an object is moving with uniform circular motion, the centripetal acceleration of the object a. is circular. b. is perpendi
never [62]
Hi Pupil Here's Your answer :::


➡➡➡➡➡➡➡➡➡➡➡➡➡


When an object is moving with uniform circular motion, the centripetal acceleration of the object \textbf{is\:Zero}.


Because the body is moving with a uniform velocity and hence, there might not be any acceleration due to it.


⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅



Hope this helps .......
5 0
3 years ago
A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.
Marizza181 [45]

Answer:

1.7×10^5 ms-1

Explanation:

From

qE= qvB

q= charge on the electron

E = electric field

v= velocity

B= magnetic field

E= vB

v= E/B= 110×10^3/0.6

v= 1.7×10^5 ms-1

3 0
3 years ago
Read 2 more answers
A 1-kg block is lifted vertically 1 m at constant speed by a boy. The work done by the boy is about:
Igoryamba

Answer:

9.8\; {\rm J}, assuming that the gravitational field strength is g = 9.8\; {\rm N \cdot kg^{-1}}.

Explanation:

Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.

By Newton's Second Law, the net force on this block would be 0. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.

Let m denote the mass of this block. It is given that m = 1\; {\rm kg}. The weight of this block would be:

\begin{aligned}\text{weight} &= m\, g \\ &= 1\; {\rm kg} \times 9.8\; {\rm N \cdot kg^{-1}} \\ &= 9.8\; {\rm N}\end{aligned}.

Hence, the force that the boy applies on this block would be upward with a magnitude of F = 9.8\; {\rm N}.

The mechanical work that a force did is equal to the product of:

  • the magnitude of the force, and
  • the displacement of the object in the direction of the force.

The displacement of this block (upward by s = 1\; {\rm m}) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:

  • the magnitude of the force that this boy exerted, F = 9.8\; {\rm N}, and
  • the displacement of this block in the direction, s = 1\; {\rm m}.

\begin{aligned}\text{work} &= F\, s \\ &= 9.8\; {\rm N} \times 1\; {\rm m} \\ &= 9.8\; {\rm J}\end{aligned}.

5 0
2 years ago
What is the best reason for having a strong hypothesis?
Elena-2011 [213]

Answer:

the answer is C

Explanation:

i did this

7 0
2 years ago
A 12000 kg boat is moving 4.25 m/s. Its engine pushes 9200 N forward, but the current pushes back at 12,500 N. How much times do
Verizon [17]

Answer:

15.5 seconds

Explanation:

Apply Newton's second law:

∑F = ma

-12500 + 9200 = (12000) a

a = -0.275 m/s²

v = at + v₀

0 = (-0.275) t + 4.25

t = 15.5 s

It takes the boat 15.5 seconds to stop.

7 0
3 years ago
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