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3 years ago

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Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible l

ight X: Gamma rays Z: Visible light Z: Gamma rays Y: Visible light Y: Gamma rays Z: Visible light

1 answer:

Marianna [84]3 years ago

4 0

The answer is Z: Gamma rays X: Visible light. Gamma ray is the shortest and has the greatest frequency and power. That is why it is in letter Z.

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How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?

Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

Q = 0.456 x 33.5 x 10^(4) = 152760 J

the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0

3 years ago

Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely

m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

Space long= 84 ft
Space wide= 42 ft
Space deep= 9 ft
Cost by yard3 = $39/yd3
Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0

1 year ago

A wind-tunnel experimentis performed on a 1/25scale model of a supersonic aircraft. The prototype aircraft flies at 450 m/s in c

KengaRu [80]

Answer: V = 504m/a

F = 4N

Explanation: please find the attached file for the solution

4 0

3 years ago

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be

ehidna [41]

Use the Inverse square law, Intensity (I)<span> of a light </span>is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>

3 0

3 years ago

What equation gives the position at a specific time for an object with constant acceleration?

malfutka [58]

Given constant acceleration, we can get the final position of an object in terms of both its initial velocity and its acceleration using one of the equations of motion.

The equation that we will use is:
Xf = Xi + Vi*t + (1/2)*a*t^2
where:
Xf is the final position of the object
Xi is the initial position of the object
Vi is the initial velocity of the object
t is the time
a is the constant given acceleration

4 0

3 years ago