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Oliga [24]
3 years ago
12

What system will break the chewed food down to a form your cell can use

Physics
1 answer:
horsena [70]3 years ago
4 0
If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients
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Someone help me , is it qualitative or quantitative.
AysviL [449]
The chart is quantitative and the beakers are qualitative.

Quantitative things always include numbers or data.
Qualitative things are based on observation.
8 0
3 years ago
A common black ant discovers a piece of bread 85 cm east of the entrance of her nest. If the ant carries 10 bits of bread back t
babunello [35]

Answer:

19 x 85 = 1,615 for distance. Displacement is 0

Explanation:

The total distance traveled by the ant in 9 round trips and one 1/2 trip, or 19 one bash way trips: 19 x 85cm = 1615cm. The displacement of the ant after the tenth trip is 0 cm ( the displacement origin is the nest.)

5 0
3 years ago
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040.
lina2011 [118]

Answer:

42.11 years old

Explanation:

Given that:

In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040

To find her age we use:

\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\

Δtm is  time interval for the observer stationary relative to the sequence of

events = 2040 - 2000 = 40 years

Δts is is the time interval for an observer moving with a speed v relative to the  sequence of event

v = velocity = 2.5 x 10^8 m/s

c = speed of light = 3 x 10^8 m/s

\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old

7 0
3 years ago
5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
4 years ago
State an advantage of using such hydraulic jack to lift a load ​
expeople1 [14]

Answer:

Explanation:

You are going to lift and press down on the 200 N many times and move only a short distance. The reward is that slowly but surely you will lift a very heavy load -- one that cannot be managed any other way but by the hydraulic jack.

5 0
3 years ago
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