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navik [9.2K]
4 years ago
11

In a newspaper, it was reported that the number of yearly robberies in Springfield in 2012 was 200, and then went down by 14% in

2013. How many robberies were there in Springfield in 2013?
SAT
1 answer:
slamgirl [31]4 years ago
6 0

Answer:

172

Explanation:

you multiply the 200 times the 14% in a format suvh as this, 200x.14 because 14% is out of 100% and you can see the 100% as 1.00, then the answer is 28 and subtract 28 from 200 ant that equals 172.

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Madame pouchets home is called.
ratelena [41]

Answer:

Villa of Green Happiness.

Explanation:

7 0
3 years ago
Newton’s law of cooling states that for a cooling substance with initial temperature t0, the temperature t(t) after t minutes ca
antoniya [11.8K]

The cooling rate of the substance is approximately 0.0732.

According to the statement, the Newton's law of cooling is defined by the following formula:

T(t) = T_{s} + (T_{o}-T_{s})\cdot e^{-k\cdot t} (1)

Where:

  • T_{s} - Final temperature, in degrees Celsius.
  • T_{o} - Initial temperature, in degrees Celsius.
  • t - Time, in minutes.
  • k - Cooling rate, in \frac{1}{min}.
  • T(t) - Current temperature, in degrees Celsius.

Please notice that substance reaches thermal equilibrium when T(t) = T_{s}, that is when temperature of the substance is equal to the temperature of surrounding air.

If we know that T_{o} = 80\,^{\circ}C, t = 15\,min, T_{s} = 50\,^{\circ}C and T(15) = 60\,^{\circ}C, then the cooling rate of the substance is:

60 = 50 + (80 - 50)\cdot e^{-15\cdot k}

\frac{60-50}{80-50}= e^{-15\cdot k}

\frac{1}{3} = e^{-15\cdot k}

k = -\frac{1}{15}\cdot \ln \frac{1}{3}

k \approx 0.0732

The cooling rate of the substance is approximately 0.0732.

To learn more on Newton's law of cooling, we kindly invite to check this verified question: brainly.com/question/13748261

7 0
2 years ago
Friction: a 50-kg box is resting on a horizontal floor. A force of 250 n directed at an angle of 30. 0° below the horizontal is
sergejj [24]

The net forces on the box parallel and perpendicular to the surface, respectively, are

∑ F[para] = (250 N) cos(-30.0°) - F[friction] = (50 kg) a

and

∑ F[perp] = F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

where a is the acceleration of the box. (We ultimately don't care what this acceleration is, though.)

To decide whether the friction here is static or kinetic, consider that when the box is at rest, the net force perpendicular to the floor is

∑ F[perp] = F[normal] - F[weight] = 0

so that, while at rest,

F[normal] = (50 kg) g = 490 N

Then with µ[s] = 0.40, the maximum magnitude of static friction would be

F[s. friction] = 0.40 (490 N) = 196 N

so that the box will begin to slide if it's pushed by a force larger than this.

The horizontal component of our pushing force is

(250 N) cos(-30.0°) ≈ 217 N

so the box will move in our case, and we will have kinetic friction with µ[k] = 0.30.

Solve the ∑ F[perp] = 0 equation for F[normal] :

F[normal] + (250 N) sin(-30.0°) - F[weight] = 0

F[normal] - 125 N - 490 N = 0

F[normal] = 615 N

Then the kinetic friction felt by the box has magnitude

F[k. friction] = 0.30 (615 N) = 184.5 N ≈ 185 N

8 0
2 years ago
Under constant conditions, the half-life of a first-order reaction.
mash [69]

Answer:

i need to know, too.. when you know tell me

8 0
2 years ago
Who intended to write an epic poem 12 books long? Question 3 options: Francis Drake Sir Walter Raleigh Edmund Spenser William Sh
Sphinxa [80]

Answer:

emund spencer

Explanation:

k12 test got it right have a great day xD

5 0
2 years ago
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