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1 month ago

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

ight in terms of the total body weight. Use the model to find a) the weight of the heart of a human whose weight is 185 lbs. Answer in units of lbs.

1 answer:

Yuliya22 [3.3K]1 month ago

8 0

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

w = human weight

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

The final weight of a human heart is 0.93 lbs.

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A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti

kicyunya [3294]

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21 day ago

Read 2 more answers

A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises

Softa [3030]

Answer:

a) The ball's velocity just prior to hitting the ground measures -6.3 m/s

b) The ball's velocity right after bouncing off the ground registers at 3.1 m/s

c) The average acceleration's magnitude is 470 m/s², and its direction is upward, forming a 90º angle with the ground.

Explanation:

To begin, let’s assess the time it takes for the ball to reach the floor:

The equation outlining the ball's position is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at given time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration triggered by gravity

We establish the ground as the reference origin.

a) Since the ball is released rather than thrown, the initial velocity v0 is 0. The direction of acceleration is downward, directed towards the origin; thus, “g” is treated as negative. When the ball contacts the ground, its position will be 0. Therefore:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The motion equation for a falling body is:

v = v0 + g * t where "v" denotes the velocity

Since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) The pebble's speed reaches 0 during its maximum height. To find the time taken for the pebble to achieve that height, we can use the velocity equation and then substitute that time in the position equation to derive the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

Replacing t in the position equation, knowing the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration can be determined by:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

The direction of the acceleration is upward, perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

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25 days ago

A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite

inna [3103]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

$v_{ey}=v_r$

We can calculate the vertical component of the explorer's swimming speed as

$v_{ey}=|v_e|cos\alpha$

Thus

$v_r=|v_e|cos\alpha$

Finding the value of $\alpha$

$\displaystyle cos\alpha=\frac{v_r}{|v_e|}$

$\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876$

Then the angle is given by

$\alpha=28.82^o$

b.

The component of the explorer's velocity that goes horizontally is

$v_{ex}=0.759sin28.82^o$

$v_{ex}=0.366\ m/s$

This represents the actual velocity directed towards the campsite

Considering that

$\displaystyle v=\frac{x}{t}$

To find t

$\displaystyle t=\frac{x}{v}$

Calculating the duration for the explorer to cross the river

$\displaystyle t=\frac{29.3}{0.366}$

$t=80\ sec$

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0

1 month ago

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

Maru [3345]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

$27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi$

Thus, the distance of the trip in miles is:

$d=16.77706389mi$

(b)

Next, we will compute the man's speed during the journey:

$v=\frac{d}{t}$

Before that, we must convert minutes to hours:

$16min*\frac{1h}{60min} =2.666666667h$

The resulting speed is:

$v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}$

Consequently:

$62.91398959\frac{mi}{h}>55\frac{mi}{h}$

Thus, it can be concluded that the driver was speeding

8 0

1 month ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

Maru [3345]

Answer:

4.05 m/s

Explanation:

We will express the varying velocities as vectors.

Newton moves northward at 3.90 m/s from Daniel's stationary position.

V_n = 3.9 j

Assuming Pauli runs relative to Daniel at velocity X.

The relative velocity of Newton as seen by Pauli will be

3.9 j - X

Given that

the relative velocity of Newton with respect to moving Pauli = 1.1 i (1.1 towards the east).

Thus,

3.9 j - X = 1.1 i

X = -1.1 i + 3.9 j.

Magnitude of X

X² = 1.1² + 3.9²

X = 4.05 m/s

Therefore, Pauli runs relative to Daniel at 4.05 m/s.

The direction will be west of north at an angle θ,

Tan θ = 1.1 / 3.9

4 0

1 month ago