The results from Rutherford's alpha particle scattering experiment led to

The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁ = 12.5 x 10⁻³ m radius.

cross sectional area = a₁

Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ = .09888 m of Hg

98.88 mm of Hg

8 0

4 years ago

A 56.0 kg box hangs from a rope. what is the tension in the rope if: the box has vy = 5.30 m/s and is speeding up at 5.10 m/s2 ?

Minchanka [31]

T = tension force in the rope in upward direction

m = mass of the box attached at end of rope = 56 kg

W = weight of the box in downward direction due to gravity

a = acceleration of the box in upward direction = 5.10 m/s²

weight of the box is given as

W = mg

inserting the values

W = (56) (9.8)

W = 548.8 N

force equation for the motion of the box is given as

T - W = ma

inserting the values

T - 548.8 = (56) (5.10)

T = 834.4 N

5 0

3 years ago

Which of the following is an example of a conductor?

nordsb [41]

Answer: B. a gold chain

7 0

3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago