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kipiarov [429]
3 years ago
10

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, an

d k > 0 is a constant. Suppose that A = A0 cos(ωt) for some constants A0 and ω. That is, the ambient temperature oscillates (for example night and day temperatures). a) Find the general solution. b) In the long term, will the initial conditions make much of a difference? Why or why not

Physics
1 answer:
atroni [7]3 years ago
3 0

Answer:

it is as shown in the attachment

Explanation:

The detailed and mathematical step by step explanation is as shown in the attachment.

The general solution x(t) = KAo/K² + w² [kcoswt + wsinwt] + Cexp(-kt)

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Answer:

The max. Energy that can be stored in the neoprene rubber capacitor will be 0.304J

Explanation:

Detailed explanation and calculation is shown in the image below

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A baseball player hits a 140 g baseball with a force of 2800 N. What is the
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B because 2800 divide by 40 is 20
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A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of ki
Neporo4naja [7]

Answer:

a = -1.961\,\frac{m}{s^{2}}, s = 16.318\,m, t = 4.079\,s

Explanation:

The equations of equilibrium for the rock are:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a

\Sigma F_{y} = N - m\cdot g = 0

After some algebraic handling, the following expression is found:

-\mu_{k}\cdot m \cdot g = m \cdot a

-\mu_{k}\cdot g = a

Deceleration experimented by the rock is:

a = - (0.2)\cdot (9.807\,\frac{m}{s^{2}} )

a = -1.961\,\frac{m}{s^{2}}

The distance travelled by the rock before stopping is:

s = \frac{(0\,\frac{m}{s} )^{2}-(8\,\frac{m}{s} )^{2}}{2\cdot (-1.961\,\frac{m}{s^{2}} )}

s = 16.318\,m

And the time is:

t = \frac{0\,\frac{m}{s}-8\,\frac{m}{s}}{(-1.961\,\frac{m}{s^{2}} )}

t = 4.079\,s

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3 years ago
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What does DRAG mean in American science?
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3 0
4 years ago
Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a to
KATRIN_1 [288]

Answer:

0.178 °C

Explanation:

Given:

The frictional force, F = 48 N

Number of rubs, n = 18

Distance covered per rub, d = 7.16 cm = 0.0716 m

Total distance covered in 18 rubs, D = 0.0716 × 18 = 1.288 m

mass of the tissues warmed, m = 0.1 kg

Now,

Work done, W = Force × Displacement

W = 48 × 1.288 = 61.8624 J

Now,

Work done is the change in heat

W = ΔQ = 61.8624 J

also,

ΔQ = mCΔT

where, C is the specific heat

for human tissue, C = 3470 J/kg°C

thus,

61.8624 J = 0.1 × 3470 × ΔT

or

ΔT = 0.178 °C

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4 years ago
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