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MatroZZZ [7]
2 years ago
14

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.
Physics
1 answer:
jekas [21]2 years ago
8 0

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is w_{h}.

We need to calculate the weight of heart of a human

Using given data

w_{h}=0.5\times w

Where, h = weight of heart

w = weight of human

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

w_{h}=0.93\ lbs

Hence, The weight of heart of a human is 0.93 lbs.

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A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f
Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

\mu mg = m\frac{v^2}{r}

where

\mu is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

v=\sqrt{\mu gr}

And by substituting the data of the problem, we find the speed at which the car begins to skid:

v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s

7 0
2 years ago
Read 2 more answers
If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?
fomenos

Answer:

Its traveling in the +x direction

Explanation:

The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.

8 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical
inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0
1 year ago
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0
1 year ago
Read 2 more answers
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