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ss7ja [257]
3 years ago
6

A street light is on top of a 8 foot pole. Joe,

Physics
1 answer:
zubka84 [21]3 years ago
5 0
8/4 = y/y-x

8y - 8x  = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps
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What happens to a blackbody radiator as it increases in temperature?
DedPeter [7]

Answer:

it gives off a range of electromagnetic radiation of shorter wavelengths.

Explanation:

I took the test

8 0
3 years ago
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Ethan pushes a wooden box across a carpeted floor. Then he pushes the same box across a smooth marble floor. Why does Ethan find
masha68 [24]

Answer:

A. The box experiences more friction on the carpeted floor

Explanation:

Friction is the force that opposes the motion of an object when it slides along a surface. The magnitude of the friction is given by

F=\mu mg

where m is the mass of the object, g is the acceleration due to gravity, and \mu is the coefficient of friction, which depends on the type of material of the surface: the larger this coefficient, the stronger the friction, the more difficult is to push the box along the surface. Generally, a smooth surface has a lower coefficient of friction, while a rough surface has a larger coefficient of friction.

In this case, Ethan find it easier to push the box on the marble floor, because marble is smoother than the carpet and so friction is weaker, while for the carpeted floor the coefficient of friction is larger and so the friction is stronger, making it more difficult to push the box.

8 0
3 years ago
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A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe
Alchen [17]
A)
2revs in 0.08s
   so in 1s thats 25revs
therefore thats <u>50π radians</u> in one second

b)
well, ω=2π/T
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and v=rω where r is in meters;
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c)
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6 0
3 years ago
What is one common source of background radiation?
boyakko [2]
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7 0
3 years ago
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You need to figure out how high it is from the 3rd floor of Keck science building to the ground. You know that the distance from
Tcecarenko [31]

Answer:1.624 m

Explanation:

Given

height of 2nd floor =5 m

time recorded is 0.58 sec

Let h be the height of third floor above 2 nd floor

h=ut+\frac{at^2}{2}

here u=0

t=time taken to cover height h

h=\frac{gt^2}{2}   ----------1

Now time taken to complete whole building length

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Subtract 1 form 2

5=\frac{g(2t+0.58)(0.58)}{2}

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1=(2t+0.58)(0.58)

t=0.57 s

Substitute the value of t in 1

h=1.624 m

4 0
3 years ago
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