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s344n2d4d5 [400]
3 years ago
7

ryan hypothesizes that darker colors heat up faster. he places a thermometer inside a red wool sock, a green cotton glove, and a

black nylon hat. whats wrong with his procedure?.
Physics
1 answer:
telo118 [61]3 years ago
3 0
Light are transfer through waves in the atmosphere and yes it  true that the darker the color is the more heat it could absorb thus it is also explain that the lighter the color is the less heat or light its absorb its because the light is bounces back through other form of light and it lessens the amount of heat in a substance. In Ryan's procedure the possible wrong that he done is the present of a green cotton glove. Green color are one of the color that bounces light and could not support the hypothesis of Ryan and the possible temperature he could get is not the accurate one.
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Answer:

a=4\ m/s^2

Explanation:

Given that,

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We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2

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The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
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Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

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Describe a vector and how it is used​
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Vectors are used in science to describe anything that has both a direction and a magnitude. They are usually drawn as pointed arrows, the length of which represents the vector's magnitude.

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4 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
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Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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Answer:

2

Explanation:

2

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