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2 years ago

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You are given $1000.00 but you must spend 1/2 the first year, and 1/2 the balance for each year thereafter. If you spend the max

imum allowed each year, at the end of what year would you have $31.25 left? How much money would you have left after 10 half-live?

1 answer:

11Alexandr11 [23.1K]2 years ago

8 0

Answer:

In this simulation, you will use small pieces of candy marked on one side. ... If you started with a sample of 600 radioactive nuclei, how many would ... were given $1000 and told that you could spend one-half of it in the first year, ... a. If you spent the maximum amount allowed in each year, at the end of what ... $31.25 left?

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3 years ago

Help:choose the correct statement plz

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Explanation:

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3 years ago

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How do you solve for a net force

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3 years ago

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the

natulia [17]

Answer:

The value is $A = 2.80 *10^{-4} \ m^2$

Explanation:

From the question we are told that

The operating temperature is $T = 2450 \ K$

The emissivity is $e = 0.350$

The power rating is $P = 200 \ W$

Generally the area is mathematically represented as

$A = \frac{P}{ e * \sigma * T^2}$

Where $\sigma$ is the Stefan Boltzmann constant with value

$\sigma = 5.67 *10^{-8} \ W/m^2\cdot K^4$

So

$A = \frac{200}{0.350 * 5.67*10^{-8} * 2450^{4}}$

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3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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2 years ago

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