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arlik [135]
4 years ago
8

Sweating and panting are examples of which characteristics of life?

Physics
2 answers:
Marianna [84]4 years ago
7 0
C.reproducing is think and if wrong I'm so very sorry
Debora [2.8K]4 years ago
3 0
B. using energy.    using alot of energy, like for exercising causes sweating and panting
You might be interested in
The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified th
hram777 [196]

Answer:

e) indicated that the speed of light is the same in all inertial reference frames.

Explanation:

In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium <em>aether</em>. They believed that even the space is not empty and filled with aether.

Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.

It thus proved that the speed of light remains same in all inertial frames.

Also, it became a base for the special theory of relativity by Einstein.

5 0
3 years ago
What determines the life cycle of a star?
Nataly_w [17]
D) the mass of the star is the correct answer.
7 0
3 years ago
A 12.0-m wire with a mass of 145 g is under tension. A transverse wave, for which the frequency is 630 Hz, the wavelength is 0.5
miskamm [114]

Answer:

The maximum traverse acceleration is   \alpha = 31346.095 m/s^2

Explanation:

Generally the maximum transverse acceleration is mathematically represented as

                   \alpha =  w^2 A

Where w is the angular velocity which is mathematically evaluated as

                 w = 2 \pi f

Substituting the values given in the question

               w = 2 * 3.142 * 630

                   = 3958.92 rad/s

  A is the A is amplitude with a value of 2.0mm = \frac{2}{1000} = 0.002m

         Substituting this into the formula for maximum transverse acceleration

                              \alpha  = 3958.96^2 * 0.002

                                  = 31346.095 m/s^2

4 0
4 years ago
Two workers pull horizontally on a heavy box but one pulls twice as hard as the other. The larger pull is directed at 25.0 west
icang [17]
1) Call  F1 the larger force and F1x and F1y its its x-and-y- components.respectively.

I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.

=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)

=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)

2) Call F2 the shorter force and F2x and F2y its components

=> cos(x) = F2x / F2 => F2x = F2*cos(x)

=> sin(x) = F2y / F2=> F2y = F2*sin(x)

3) You know that:

- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0

a)  F1x + F2x = 0

=>  -F1*cos(65) + F2*sin(x) = 0

=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)

Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524

=> x = arcsin(0.84524) =  57.7


b) F1y + F2y = 430 =>

F1 sin(65) + F2*sin(57.7) = 430 =>

0.9060F1 + 0.84524F2 430

F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430

=> F2 = 430 / 1.7512 = 245.54 N

=> F1 = 2*245.54 =491.1N

There you have the two forces.

The angle of the shorter force is 57.7 measured from the east to the north (this is north of east),  which would be 90 - 57.7 = 32.3 degrees east of north..

 Then the shorter force is 245.5 N at 32.3 degrees east of north

And the larger force is 491.1 N at 25.0 degrees west of north.

 
3 0
4 years ago
a summer speeds up from 1.1m/s to 1.3m/s during the last 20 s of the workout. what is the acceleration during this interval
Nataly_w [17]

Answer:

acceleration= 0.01 m/s²

Explanation:

8 0
3 years ago
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