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BigorU [14]
3 years ago
14

According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f

rom that of T 1, the body will either cool or warm to temperature T(t) after t minutes, where:
T(t) = T 0 + (T 1 - T 0)e kt

and k is a constant.

A cup of coffee with temperature 140°F is placed in a freezer with temperature 0°F. The constant k ≈ -0.0815. Use Newton's Law of Cooling to find the coffee's temperature, to the nearest degree Fahrenheit, after 15 minutes.

The temperature is about
degrees Fahrenheit.
Physics
2 answers:
Mila [183]3 years ago
3 0

We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.

T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

Anon25 [30]3 years ago
3 0

Answer:

-34º

Explanation:

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INCREASE in temperature of the material practically increase the energy of the particles. which increases their motion due to increase in energy . thus when the temperature is decreased the energy level decreases which causes the particle's motion to slow down.. the motion of the particle is highly reduced when the temperature is lowered
4 0
3 years ago
Can any one answer these two
dalvyx [7]

) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer.

1) (2.68 x 10¯5) x (4.40 x 10¯8)

The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures gives 1.18 x 10¯12 as the answer.

2) (2.95 x 107) ÷ (6.28 x 1015)

The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives 4.70 x 10¯9 as the answer.

3) (8.41 x 106) x (5.02 x 1012)

The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives 4.22 x 1019 as the answer.

When done as a division, the answer to this problem is 1.68 x 10¯6.

4) (9.21 x 10¯4) ÷ (7.60 x 105)

The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives 1.21 x 10¯9 as the answer

7 0
2 years ago
A team of bicyclists are on bikes that require 512 J of work to ride. It takes 432 J of work for the bike to turn the gears. Wha
Alja [10]

Answer:

84.4 %

Explanation:

Mechanical efficiency = output work/input work × 100 %

output work = 432 J of work for the bike to turn the gears

input work = 512 J of work to ride.

Mechanical efficiency =  432 J/512 J × 100 %

= 0.844 × 100%

= 84.4 %

8 0
3 years ago
How does the law of conservation apply to Earth's Energy Budget?
Molodets [167]

Answer:

The balance between incoming energy from the sun and outgoing energy from Earth ultimately drives our climate. This energy balance is governed by the first law of thermodynamics, also known as the law of conservation of energy.

4 0
2 years ago
A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str
ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0
3 years ago
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