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3 years ago

When sunlight interacts with a raindrop to produce a rainbow, how many refractive events occur for each drop

Physics

1 answer:

erastova [34]3 years ago

6 0

Answer:

The correct answer will be "2".

Explanation:

The mechanism of rainbow formation starts whenever sunlight is shone on a raindrop. When the sun's light waves hit and go through a droplet of water, their pace changes a little. This induces bending or "refracting" of the direction of light.
Light moves in such a single direction, unless someone represents something, refracts, or diffracts everything that. When some of these things have happened, the balanced information of color is isolated, in which each could still be seen.

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Technician A says that battery corrosion is commonly caused by spilled electrolyte. Technician B says that battery corrosion is

klio [65]

Answer:

B is correct

Explanation:

Battery contains sulfuric acid H2SO4. As the battery cools and eats with the running of the motor, hydrogen gasses is released from electrolytic reactions and escapes through the vents. it mixes with other particulates. The mixtures react at the electrical terminals of the battery and corrosion is formed.

Corrosion can be prevented by the addition of anti corrosion gel. It can also be removed by scrubbing the terminals with a hard brush

3 0

4 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago

Whích phase change is the result of particles moving more quickly? (2 points)

sdas [7]

Answer:

liquid to gas

Explanation:

let me know if this helps

8 0

3 years ago

Please answer help me

Andrews [41]

I believe the answer is x

7 0

3 years ago

suppose 1 gold bar is 6 cm long, 2 cm high, and 3 cm wide. how many whole gold bars will you be able to fit in a vault with a vo

Gemiola [76]

To determine the number of gold bars that would fit in the vault, determine the volume (v) of a single gold bar.
v = l x h x w
where l, h, and w are length, height and width, respectively. Substituting the known values,
v = (6 cm)(2 cm)(3 cm) = 36 cm³
The number of gold bars can be determined by dividing the volume of the vault by the volume of the gold bar,
n = (500 cm³ / 36 cm³) = 13.89
Thus, the maximum number of gold bars in the vault is 13.

5 0

3 years ago