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Aleksandr
18 days ago
12

When you hold your hands at your sides, you may have noticed that the veins sometimes bulge--the height difference between your

heart and your hands produces increased pressure in the veins. The same thing happens in the arteries. If the average arterial pressure at your heart is a typical 100 mmHg, what is the average arterial pressure in your hands when they are held at your side? Assume your hands are 60 cm below your heart.
Physics
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In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c
Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

GDP=\pi } \frac{d^{2} }{4}

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

GDP_{b} =\pi \frac{3^{2} }{4}

Conversely, in darkness, the diameter typically enlarges to 7 mm.

GDP_{b} =\pi \frac{7^{2} }{4}

This indicates an increase in light-gathering capacity.

Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}

Thus, the amount of light the eye can capture is 31.4 mm².

3 0
3 months ago
An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
Ostrovityanka [3204]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time t_{sound}. Thus, we can derive:

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

t_{arrow} + t _{sound} = 1 s.

Using this relationship in the distance formula for sound allows us to write:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Substituting the value of d from the first equation yields:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, after some calculations, we can proceed further:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Finally, the value is inserted into the initial equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
3 months ago
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
serg [3582]

Response:

(a) W=-19.25J

(b) W=-52.8J

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

W=nRTln(\frac{V_1}{V_2} )

The moles are calculated based on the provided mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Consequently, the work amounts to:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Best regards.

4 0
2 months ago
An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s2 and there is no air. To d
Keith_Richards [3271]

Answer:

12.1 seconds

Explanation:

t = time duration

u = initial speed

v = final speed = 0

s = distance = 120 m

a = lunar gravity acceleration = 1.67 m/s²

Motion equation

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -1.67\times 120-0^2\\\Rightarrow u=\sqrt{2\times 1.67\times 120}\\\Rightarrow u=20.02\ m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20.2}{-1.67}\\\Rightarrow t=12.1\ s

The rock takes 12.1 seconds to reach the bottom of the crater.

5 0
3 months ago
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