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sertanlavr [38]
3 years ago
9

How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

minus initial position D. initial position minus final position
Physics
1 answer:
e-lub [12.9K]3 years ago
5 0
The answer is C. Final position minus initial position.
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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
The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane
umka21 [38]

Answer:

\Delta p_x=7.75\times 10^{-25}\ kg-m/s

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, \Delta x=68\ pm=68\times 10^{-12}\ m

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }

Putting all the values, we get :

\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s

So, the momentum component of this electrons is greater than 7.75\times 10^{-25}\ kg-m/s.

6 0
4 years ago
Personality theorists often focus on ________________.
raketka [301]

Personality theorists often focus on differences in individuals.

Answer: Option A

<u>Explanation: </u>

According to the personality psychology, personality scientists study personality and its variations amongst the individuals. They often focus on the thoughts, judgements and social behaviour of the individuals that they exhibit over time. With these observations, they form a report which better conclude their mental and emotional state.

By studying different types of personalities and respective behaviour, personality theorists are able to predict the diversity in human nature and that whether a person is mentally and emotionally healthy. If not, theorists conduct therapies to get them in order and hence, lead a more meaningful and healthy life.

8 0
4 years ago
In a glider stunt at an air show, a towing airplane (motorized plane pulling the gliders) takes off from a level runway with two
irakobra [83]

Answer:

minimum length of runway is needed for take off 243.16 m

Explanation:

Given the data in the question;

mass of glider = 700 kg

Resisting force = 3700 N one one glider

Total resisting force on both glider  = 2 × 3700 N = 7400 N

maximum allowed tension = 12000 N

from the image below, as we consider both gliders as a system

Equation force in x-direction

2ma = T -f

a = T-f / 2m

we substitute

a = (12000 - 7400 ) / (2 × 700 )

a = 4600/1400

a = 3.29 m/s²

Now, let Vf be the final speed and Ui = 0 ( as starts from rest )

Vf² = Ui² + 2as

solve for s

Vf² = 0 + 2as

2as = Vf²

s = Vf² / 2a

given that take of speed for the gliders and the plane is 40 m/s

we substitute

s = (40)² / 2×3.29

s = 1600 / 6.58

s = 243.16 m

Therefore, minimum length of runway is needed for take off 243.16 m

4 0
3 years ago
A bullet is shot from a rifle with a velocity of 720 m/s. What is the velocity of the bullet in km/h.
Elan Coil [88]

Answer:

2592 km/h

Explanation:

Given that a bullet is shot from a rifle with a velocity of 720 m/s. What is the velocity of the bullet in km/h.

The velocity = 720 m/s

Solution

To convert metres per second to kilometer per hour, you will multiply by 3600 and divide by 1000

720 × 3600/1000

720 × 3.6

72 × 36

2592 km/h

Therefore, the velocity of the bullet in kilometer per hour is 2592 km/h

8 0
3 years ago
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