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lutik1710 [3]
3 years ago
5

If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dr/dt. dA dt = dr dt (b

) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 25 m?
Physics
2 answers:
LenKa [72]3 years ago
6 0

Answer:

A) dA/dt = 2πr(dr/dt)

B) dA/dt = 50π m²/s

Explanation:

A) The formula for Area is;

A = πr²

Since, we want to find dA/dt. Thus,

(A)(1/dt) = (πr²)(1/dt)

Thus, differentiating;

(dA/dr)(1/dt) = 2πr(1/dt)

Multiply both sides by dr to obtain;

dA/dt = 2πr(dr/dt)

(b) we want to find rate of area (dA/dt) when r = 25m and dr/dt = 1 m/s

since we know that, dA/dt = 2πr(dr/dt), we can solve it as;

dA/dt = 2π(25)(1)

dA/dt = 50π m²/s

Lorico [155]3 years ago
5 0

Answer:157.1 m^2/s

Explanation:

Given

A is the area of circle with radius r

A=\pi r^2

Differentiate w.r.t time

\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times \frac{\mathrm{d} r}{\mathrm{d} t}

Also \frac{\mathrm{d} r}{\mathrm{d} t}=1 m/s at r=25 m

\frac{\mathrm{d} A}{\mathrm{d} t}=\pi \times 2\times 25\times 1=157.1 m^2/s

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Yes

Explanation:

An object can be moving (have kinetic energy) and be elevated above the ground at the same time (and also have potential energy).

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A machine which has an energy loss of 10% will have efficiency of​
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90%

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The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can
Sonja [21]

Answer:

Explanation:

a ) speed of passenger = circumference / time

= 2π R / Time

= 2 x 3.14 x 50 / 60

= 5.23 m /s

b )

centrifugal force = m v² /R

= (882 /9.8 ) x 5.23² / 50

= 77.47 N

Apparent weight at the highest point

real weight  - centrifugal force

= 882 - 77.47

= 804.53 N

Apparent weight at the lowest point

real weight  + centrifugal force

= 882 +77.47

= 959.47 N

c )  if the passenger’s apparent weight at the highest point were zero

centrifugal force = weight

mv² /R = mg

v² = gR

= 9.8 X 50

v = 22.13 m /s

d )

apparent weight

mg - mv² / R

= 882 - (882 / 9.8 )x 22.13²/50

= 882 +  882

= 1764 N  

=

6 0
3 years ago
platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin
posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

a=g=9.8 m/s^2 is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s

For the diver jumping from 10 m, s = 10 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

s=ut+\frac{1}{2}at^2

And since u = 0, it can be reduced to

s=\frac{1}{2}at^2

For the diver jumping from 5 m, s = 5 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s

For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

5 0
3 years ago
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