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koban [17]
2 years ago
8

A loaf of bread is removed from an oven at 350° F and cooled in a room whose temperature is 70° F. If the bread cools to 210° F

in 20 minutes, how much longer will it take the bread to cool to 185° F.
Physics
1 answer:
jasenka [17]2 years ago
7 0

Answer:

The bread will take 216.40 minutes to cool to 185°F

Explanation: Newton law of cooling

states that the rate of cooling of an object is inversely proportional to the difference of temperatures between the object and its surroundings i.e. dTdt=−kT, where t is the time taken and T is the difference of the temperatures between the object and its surroundings.

This gives us T as a function of t and is given by T(t)=ce−kt.

T(0)= Ce^-kt

C=35₩F- 185F= 165F

T(20) = 165×e^-20k=(350-70)

T(20)= e^-20k =280/165

-20k= ln(280/165)

-20k= ln(1.697)

-20k= 0.5289

K= 0.5289/20

K= 0.0264

If it cools at 185F in t minutes

165e^-0.0264×t=0.5

e^-0.0264t= 0.5/165

-0.0264t= on 3.03×10^-3

-0.0264t= -5.7129

t= 5.7129/0.0264

t=216.40minutes

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ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

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8 0
3 years ago
An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False
Umnica [9.8K]
Hi There! :)


An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P
7 0
3 years ago
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(b) Two containers made of insulating material contain the same volume of water at room
masha68 [24]

Answer:

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5 0
2 years ago
an object is producing a sound that has a wavelength in air of 2.69m. If the speed of sound in air is 346m/s, what is the freque
Misha Larkins [42]

Answer:

129.74 Hz

Explanation:

Given:

Wave velocity ( v ) = 346 m / sec

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We have to calculate Frequency ( f ) :

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v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 346 / 2.69 Hz

= > f = 34600 / 269 Hz

= > f = 129.74 Hz

Hence, frequency of sound is 129.74 Hz.

5 0
2 years ago
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