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4 years ago

Singularity is difficult to understand due to the fact that the idea of space and time are not applicable given the definition.

1 answer:

5 0

I assume you are talking about the space-time singularity.

Yes, the gravitational singularity is absurd since the gravitational field is infinite so that the celestial body does NOT depend on the coordinates anymore, i.e. the normal space-time theorem does NOT apply anymore.

However, this can be predicted by general relativity, where the singularity can appear at the center of the black hole.

So if you are talking about classical physics, yes it's difficult. But if you are talking about modern physics, then it's predictable and analyzible.

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The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

4 0

3 years ago

A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj

Alika [10]

Answer:

265 J

Explanation:

$Energy=PE+KE=mgh+ 0.5mv^{2}$ where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

$Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J$

4 0

3 years ago

Just these 2 question. I have no idea what it's asking..

Murljashka [212]

Answer: did you get the answers?

Explanation:

5 0

3 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

4 years ago

1. A ball is thrown downward with an initial speed of 22 m/s on Earth. a. What is the acceleration of the ball? b. Calculate the

vitfil [10]

Answer:

Explanation:

The acceleration of the ball would be due to the downward force of gravity, 9.8m/s^2. In order to find the displacement given that interval of time, you have to use the corresponding kinematic formula:

$d=v_it+at^2/2$