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Ghella [55]
3 years ago
14

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

mperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 185 degrees Fahrenheit when freshly poured, and 3 minutes later has cooled to 172 degrees in a room at 78 degrees, determine when the coffee reaches a temperature of 147 degrees.
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

6.77 minutes

Explanation:

172 degree - 78 degree = (185 degree - 78 degree)e−2 k

=> 94 = 107

e−2 k => 94 ÷ 107

k => ln (94÷107) / 2

147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]

=> 69 = 107 e^ [ln (94÷107) / 2]

e^[ln (94÷107) / 2] =69 ÷ 107

=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]

t=> -0.4387 ÷ -0.0648

t => 6.77 minutes.

Therefore, the final answer to the question is 6.77 minutes.

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Answer:The Answer Is D

Explanation:

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All of the following processes are the ones that lead directly to the formation of sedimentary rock EXCEPT-
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Answer:

melting of rock deep underground.

Explanation:

The melting of rocks deep underground does not produce sedimentary rocks. Most igneous rocks are produced by this process.

When molten rocks underground called magma is solidified in the subsurface, it results into the formation of igneous bodies.

  • Sedimentary rocks forms by the accumulation of sediments.
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4 0
2 years ago
A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass
makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s

Velocity of cart with child on top is 3.38 m/s

7 0
3 years ago
Can you respond this two questions, please? :
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Where's the diagram for question 1?

6 0
3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
  • =18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
3 years ago
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