PHYSICS COLLEGE

The left ventricle of a resting adult's heart pumps blood at a flow rate of 85.0 cm3/s, increasing its pressure by 110 mm Hg, its velocity from zero to 25.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output (in W) of the left ventricle. Note that most of the power is used to increase blood pressure.

Answers

Answer 1

Answer:

Answer:

P = 1.29625 W

Explanation:

Given

Q = 85.0 cm ³ / s, p₁ = 110 mmHg, u₁ = 25.0 cm / s, h = 5.0 cm

Also knowing the density of the blood is

ρₐ = 1.05 x 10 ³ kg / m³

Δp₁ = 110 mmHg * 133.322 Pa / 1 mmHg

Q = 85.0 cm³ / s = 85.0 x 10 ⁻⁶ m³ / s

To calculated the power

P = H * Q

H = Δp₁ + ¹/₂ * ρₐ * ( u₁² - v₂²) + ρₐ * g *Δh

H = 14.666 x 10 ³ Pa + 0.5 * 1.05 x 10 ³ kg / m³ * ( 25 x 10 ² m /s )² + 1.05 x 10 ³ kg / m³ * 9.8 m /s² * 0.05 m

H = 15.25 x 10 ³ Pa

P = 15.25 x 10 ³ Pa * 85.0 x 10 ⁻⁶ m³ / s

P = 1.29625 W

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1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building? 2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answers

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1. $s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m$

2. $v = gt = 9.81*3.3 = 32.373 m/s$

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, $v_0$ is the initial velocity, $\Delta s = 48.415$ is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

$0 - v_0^2 = 2*(-9.81)*48.415$

$v_0^2 = 950$

$v_0 = √(950) = 30.82 m/s$

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = (7.4)/(cos(70))$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

When dots are further apart on a ticker-tape diagram, it indicates an object is moving

Answers

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Hope this helps!

Answer:

At a higher velocity.

A wire is wrapped around a piece of iron, and then electricity is run through the wire. What happens to the iron?

Answers

Search ResultsBy simply wrapping wire that has an electrical current running through it around a nail, you can make an electromagnet. When the electric current moves through a wire, it makes a magnetic field. ... You can make a temporary magnet by stroking apiece of iron or steel (such as a needle) along with a permanent magnet.

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A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?

Answers

Answer:

the lowest frequency is $7.5* 10^(14) Hz$

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν= $(v)/(L)$

frequency ν be lopwest when L will be heighest

ν(lowest)= $(3* 10^8)/(400* 10^-9)$

ν= $7.5* 10^(14) Hz$

The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.5 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor?