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3 years ago

10

A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At

what speed is the tip of Joe’s shadow moving from the base of the pole when he is 10 feet from the pole?

1 answer:

Gekata [30.6K]3 years ago

6 0

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

$\frac{x'}{x+x'}=\frac{3}{9}$

3x'=x+x'

x=2x'

and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

Differentiating

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}$

$4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}$

$\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s$

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8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

3 years ago

Potassium is a crucial element for the healthy operation of the human

Degger [83]

Answer:

1

The mass of the Potassium-40 is $m_{40}} = 2.88*10^{-6} kg$

2

The Dose per year in Sieverts is $Dose_s = 26.4 *10^{-10}$

Explanation:

From the question we are told that

The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

Their abundance is 93.26%, 0.012% and 6.728%

The mass of potassium contained in human body is $m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg$ per kg of the body

The mass of the first body is $m_1 = 80 \ kg$

Now the mass of potassium in this body is mathematically evaluated as

$m_p = m * m_1$

substituting value

$m_p = 80 * 0.0003$

$m_p =0.024 kg$

The amount of Potassium-40 present is mathematically evaluated as

$m_{40}} =$ 0.012% * 0.024

$m_{40}} = \frac{0.012}{100} * 0.024$

$m_{40}} = 2.88*10^{-6} kg$

The dose of energy absorbed per year is mathematically represented as

$Dose = \frac{E}{m_1}$

Where E is the energy absorbed which is given as $E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}$

Substituting value

$Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}$

$Dose = 22*10^{-10} J/kg$

The Dose in Sieverts is evaluated as

$Dose_s = REB * Dose$

$Dose_s = 1.2 * 22*10^{-10}$

$Dose_s = 26.4 *10^{-10}$

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3 years ago

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Travka [436]

Answer:

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Explanation:

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ale4655 [162]

Answer:

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Explanation:

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3 years ago

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and

Liono4ka [1.6K]

a) $5 \Omega$ , 1.6 A

b) $6 \Omega$ , 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0

3 years ago