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Klio2033 [76]
3 years ago
12

Andre is studying snails and how fast they travel. He observes three groups of snails containing four snails each. Here are the

recorded observations from Andre’s notebook: Group A Snails averaged a speed of 4.7cm/min in a petri dish. Group B Snails averaged a speed of 2.8 cm/min on sandpaper. Group C Snails averaged a speed of 3.3 cm/min in the dirt. Based on this data, what is affecting the snails’ speed?
Physics
1 answer:
Olin [163]3 years ago
4 0
The material that the snails are moving on affects their speed. On a smooth material, (petri dish) the snails moved the fastest. On a rough material (sandpaper) the snails moved the slowest. On dirt, a compromise between smooth and rough, the snails moved at a medium pace. The material and possibly friction affect this.
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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
2 years ago
How long does it take an airplane to fly 5000 m if it maintains a speed of 240 meters per second?
kolbaska11 [484]

Answer:

20 5/6 sec

Explanation:

To find the solution we divide 5000 by 240

However, when you see a problem, always try to simplify

5000/240=500/24=250/12=125/6

Now the division is much easier

20 5/6 sec

8 0
2 years ago
What would happen if you place 2 Plane Mirrors parallel to each other?
e-lub [12.9K]

Answer:

No image will be observed.

Explanation:

Images that are created by mirrors are virtual images. This virtual image can only be seen by an observer. In this case, an infinite number of images or no image will be created here as both will be reflecting their own images. Light will continuously bounce back and forth reflecting the same image.

6 0
3 years ago
Two forces F1 and F2 are acting on a block of mass m=1.5 kg. The magnitude of
nika2105 [10]

Answer:

Explanation:

Component of force F₁ in right direction = F₁cos37

= 12 x cos37 = 9.58 N .

Component of force F₂ in  vertically upward direction = F₁sin37

= 12 x sin37 = 7.22 N .

a ) Let normal force be R

R + F₁sin37  = mg

R + 7.22 = 1.5 x 9.8 = 14.7

R = 7.48 N .

b )

Net force in horizontal direction = F₂ - F₁cos37

= F₂ - 9.58

This is equal to zero as body is moving with zero acceleration

F₂ - 9.58 = 0

F₂ = 9.58 N

c ) If body is moving with acceleration of 2.5 m /s² along the direction of F₂

F₂ - 9.58 = 1.5 x 2.5 = 3.75

F₂ = 13.33 N .

6 0
2 years ago
Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he p
lapo4ka [179]

Answer:

He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.

6 0
2 years ago
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