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lord [1]
3 years ago
13

The sense of smell helps protect us against danger. Explain why.

Physics
1 answer:
atroni [7]3 years ago
6 0

Well we can always smell things like if food smells bad or if we smell something we have to know whether that smell is good or bad. You could be able to smell animals that are dangerous. Or animal droppings, to know what animal it is if you are in danger.

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Graph the following data tables on different graphs.
Anna [14]

Answer:

Sjjsjsjsjsjsjsjwjwjw

8 0
2 years ago
Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib
Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

6 0
2 years ago
In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (
omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0
3 years ago
The critical angle for water is 49°. If a ray of light
Sonja [21]

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49      Ni = 1.325 deg     the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg  

So refraction can occur  outside the pond and you do not have total internal refection.

 

3 0
3 years ago
Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
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