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VMariaS [17]
4 years ago
8

(a) How long will it take an investment to double in value if the interest rate is 8% compounded continuously? (Round your answe

r to two decimal places.) yr (b) What is the equivalent annual interest rate? (Round your answer to two decimal places.)
Physics
1 answer:
NeX [460]4 years ago
5 0

Answer:

Explanation:

Given

rate of interest= 8%

For continuously compounded interest

A=Pe^{rt}

Where A=amount

P=Principal

r=rate of interest

t=time

2P=Pe^{0.08t}

2=e^{0.08t}

taking log both sides

\ln 2=0.08 t

t=\frac{\ln 2}{0.08}

t=8.66 yr

(b)Equivalent annual interest

2=(1+i)^t

2=(1+i)^{8.66}

2^{0.1154}=1+i

i+1=1.0832

i=0.0832

8.32 %

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PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph
hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

x = \frac{L}{\frac{W}{F} +1}

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8  = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W  = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8  = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

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3 years ago
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A bug of mass 2.2 g is sitting at the edge of a cd of radius 3.0 cm. if the cd is spinning at 280 rpm, what is the angular momen
m_a_m_a [10]
M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.

The angular speed is
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The moment of inertia of the bug is
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Calculate the angular momentum of the bug.
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Answer: 5.806 x 10⁻⁵ (kg-m²)/s

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4 years ago
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