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3 years ago

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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.

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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What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Stolb23 [73]

Answer:

$1.1\cdot 10^{-10}C$

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

$q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C$

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If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

Gre4nikov [31]

Answer: 500 s

Explanation:

Speed $v$ is defined as a relation between the distance $d$ and time $t$ :

$v=\frac{d}{t}$

Where:

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$d=1.5(10)^{11}m$ is the distance between the Earth and Sun

$t$ is the time it takes to the light to travel the distance $d$

Isolating $t$ :

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$t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}$

Finally:

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An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

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where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

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Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

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Learn more about dynamic pressure here:

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mrs_skeptik [129]

Answer:

5070

Explanation:

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Read 2 more answers

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

Put the value into the formula

$I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2$

$I=1.97=2.0\ A$

Hence, The current is 2.0 A.

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