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andreev551 [17]
3 years ago
7

Hello:) how to do 3?:)

Physics
1 answer:
Alecsey [184]3 years ago
4 0

Answer:

A

Explanation:

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Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit
pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let T_{0} = 70^{0}F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

\frac{DT}{dt} = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

\frac{DT}{t-70} = <em>kdt</em>

In [T-70]= <em>kt </em>+C_{1}

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Finding he value for C_{2} using the initial value of T (0)= 300, therefore we get:

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T= 70+ 230 e^{kt}

Finding the value for <em>k </em>using T (3)  = 200, therefore we get

T (3) = 200

e^{3k} = \frac{13}{23}

<em>K </em>= \frac{1}{3} in \frac{13}{23}

= -0.19018

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T(t) = 70+230e^{-0.19018}

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3 years ago
In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo
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Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

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Using equation of motion

s=ut+\dfrac{1}{2}at^2

Where, s = distance

t = time

a = acceleration

Put the value into the formula

60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2

a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}

a=0.0244\ m/s^2

Hence, The acceleration of the both masses is 0.0244 m/s².

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In which of the following examples does the object have both kinetic and potential energy? Select all that apply.
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