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Aliun [14]
4 years ago
8

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 5.5 mi2/hr. How rapidly is rad

ius of the spill increasing when the area is 6 mi2?
Physics
1 answer:
Travka [436]4 years ago
8 0
<h2>The radius will increase at the rate of 0.64 mi/hr </h2>

Explanation:

The area of a circle can be represented by  A = π r²            I

Differentiating both sides w.r.t time

\frac{dA}{dt} = 2π r \frac{dr}{dt}                                    II

Dividing II by I , we have

\frac{dA}{A} = 2 x  \frac{dr}{r}

substituting the values

\frac{dr}{r} = \frac{5.5}{12} = 0.46 mi per unit radius

or dr = 1.4 x 0.46 = 0.64 mi/hr

here 1.4 mi is the radius , when area of circle is 6 mi²

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Read 2 more answers
An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
4 years ago
A light ray incident from medium 1 to medium 2, where n1&gt;n2. When the incident angle exceed the critical angle ac, the refrac
vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence

{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

{\theta_{critical}} is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0
4 years ago
What is the heat extracted from the cold reservoir for the refrigerator?
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What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J . 

<span>Qc= _________ </span>

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K=105J
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4 years ago
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