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aliina [53]
3 years ago
13

Des Linden won the Boston marathon in 2018, becoming the first American woman to win since 1985. The harsh conditions (heavy rai

n and a headwind) led to a winning time of 2 hrs 39 minutes and 54 seconds. If she weighed 99 pounds, her average heart rate during the race was 170 beats per minute and her heart pumped 2.5 mLs of blood per kg body mass per beat, how much blood did her heart pump during the race?
Physics
1 answer:
ankoles [38]3 years ago
6 0

Answer:

3050.6 Litre .

Explanation:

Total time of heart beat = Total time of race  = 2 hrs , 39 minutes and 54 seconds

= 2 x 60 + 39 + 54/60 min

= 120 + 39 + .9 min

= 159.9 min

rate of heart beat = 170 per min

Total no of heart beat during race = 170 x 159.9

volume of blood per kg per beat = 2.5 mL per kg of weight

body weight  = 99 pounds = .4535 x 99  kg = 44.89 kg

volume of blood  per beat = 2.5 mL x 44.89 mL

= 112.225 mL .

Total required volume of blood =  112.225  x 170 x 159.9 mL

= 3050612 mL

= 3050.6 L.

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What is the refraction index of a medium?
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A 1.90-kg mass vibrating up and down on the end of a vertical spring has a maximum speed of 2.30 m/s. What is the total potentia
Pepsi [2]

Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg

Therefore the maximum kinetic energy of motion is given by

Kinetic Energy, KE = \frac{1}{2} mv^{2}

Where,

m = Attached vibrating mass = 1.90 kg

v = velocity of the string = 2.3 m/s

Therefore Kinetic Energy, KE = \frac{1}{2}×1.9×2.3² = 5.0255 J

From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion

Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion

That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

Total PE = Maximum KE = 5.0255 J.

6 0
3 years ago
An archer pulls a bowstring back a distance of 20 cm with an average force of 75 N. The arrow has a mass of 20.0 g. When he rele
Masja [62]

To solve this problem we will apply the concepts related to the work theorem for which it is defined as the product of Force and distance. In turn, we will use the energy conservation theorem for which the applied work must be equivalent to the total kinetic energy on the body.

The work is defined as

W = Fd

Here,

F = Force

d = Displacement

Replacing with our values we have that

W = 75*0.2

W = 15J

Now by conservation of energy,

W = KE

W = \frac{1}{2}mv^2

15 = \frac{1}{2} (20*10^{-3}kg)(v^2)

Solving for v,

v = 10\sqrt{15}

v = 38.72 \approx 39

Therefore the correct answer is D.

8 0
4 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
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