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Iteru [2.4K]
3 years ago
11

Which two parts do the capillaries surround?

Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

most probably it's C

Explanation:

alveoli is surrounded by capillaries and whit that option small intestines is also there

Elis [28]3 years ago
6 0

Answer:

do we have to choose 2 answers here?

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PLEASE HELP!!! 12-14!!!!!
kondor19780726 [428]
12-14 equals -2.....
8 0
3 years ago
Why there is tides in sea and ocean
AleksandrR [38]

Answer:

<em>    Tides are the rise and fall of sea levels. Tides occur due to the gravitational forces exerted on the Earth by moon, and to a lesser extent, the sun. That's why tides occur in ocean or sea.</em>

Hope this helps

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7 0
3 years ago
How will the speed of a sound wave change if the temperature of the medium rises?. . Select one of the options below as your ans
Stolb23 [73]
The correct answer for this question is " A.. The speed will i<span>ncrease." The higher the temperature, the more energy it has, and making it to vibrate faster. Since it vibrates faster, then sound waves can travel at a higher speed. In other words, it will make the the speed to increase.</span>
5 0
3 years ago
Read 2 more answers
When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.
Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

F=mg=(4.32)(9.8) = 42.3 N

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

x=2.92 cm = 0.0292 m

We can find the spring constant by using Hooke's law:

F=kx

where k is the spring constant. Solving for k,

k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

F'=m'g=(1.50)(9.8)=14.7 N

So, if we use Hooke's law again, we can find the new stretching of the spring:

x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

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

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J

6 0
3 years ago
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate
LenKa [72]

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

6 0
3 years ago
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