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USPshnik [31]
3 years ago
12

By how much will the length of a chicago concrete walkway that is 18 m long contract when the equipment drops from 24 degrees ce

lcius in July to (-16 degrees celcius) in Janruary?
Physics
1 answer:
xenn [34]3 years ago
4 0

Answer:

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

Explanation:

Thermal expansion or compression is given by ΔL = LαΔT

Here Length of Chicago concrete walkway, L = 18 m

         Change in temperature, ΔT = (-16 - 24) = -40 °C

         Coefficient of thermal expansion for concrete, α = 12 x 10⁻⁶ °C⁻¹

Substituting

    ΔL = LαΔT = 18 x 12 x 10⁻⁶ x (-40) = -8.64 x 10⁻³ m

Contraction of Chicago concrete walkway = 8.64 x 10⁻³ m

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A(n) _________________ is a push or pull that one object exerts on another
Naya [18.7K]

Answer:

Explanation

     it is a force

5 0
3 years ago
What is evaporation? Condensation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences.
krok68 [10]

Answer:

Evaporation is the physical change in which a substance converts from its <u>liquid</u> state to its <u>gaseous</u> state. Condensation is the physical change in which a substance converts from its <u>gaseous</u> state to its <u>liquid</u> state.

Explanation:

Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.

7 0
4 years ago
A sphere is charged with electrons to -6 x 10^-6C. How many electrons make up this charge? The elemental charge is 1.6 x 10^-19
just olya [345]

Answer:

3.75*10^{-13}  electrons

Explanation:

The total charge Q is the sum of the charge of the N electrons contained in the sphere:

Q=N*q_{e}

q_{e}=-1.6*10^{-19}C    charge of a electron

We solve to find N:

N=\frac{Q}{q_{e}}=\frac{-6 x 10^{-6}}{-1.6 x 10^{-19}}=3.75*10^{-13}

7 0
4 years ago
A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
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7 0
3 years ago
Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain
algol13

Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

The initial mass of the sample = 1 microgram

After 5 half-lives, the mass should reduce to 1/32 of the original.

So the concentration left = 1/32 of 1 microgram = 1/32 microgram

7 0
3 years ago
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