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3 years ago

14

A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

n elevation of 6100 feet. What is the length of the trail (to the nearest foot)

1 answer:

9966 [12]3 years ago

7 0

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

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An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc

Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

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

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

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4 years ago

PLEASE HELP Due today!

BigorU [14]

So i believe is exercise:)

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3 years ago

Muitos remédios são tomados em doses menores que o mg. Um comprimido de certo remédio tem 0,025 mg de uma certa substância. Com

Sergio039 [100]

Answer:

Explanation:

<u>The question is asked in portuguese</u>

<u>Hence, it will be answered in portuguese</u>

1kg = 1000g

e

1 g = 1000mg

então

1 kg = 1 x (1000 x 1000)

1 kg = 1x 1.000.000

1 kg = 1.000.000mg

assim

1 comprimido -------------> 0,025 mg

x comprimido ------------> 1.000.000 mg ( só cruzar)

0,025(x) = 1(1.000.000)

0,025x = 1.000.000

x = 1.000.000/0,025

x = 40.000.000

<h3>1kg são feitos 40.000.000( Quarenta milhões de comprimidos)</h3>

7 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

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3 years ago

Consider the two electron arrangements for neutral atoms a and b. Are atoms a and b the same element? a - 1s2, 2s2, 2p6, 3s1 b -

Agata [3.3K]

The two neutral atoms A and B have the same number of electrons and atomic number 11. So, the two elements are said to be same.

The electronic configuration of the element is the arrangement of the electrons in the atom of the element in energy levels, orbitals around the nucleus.

The electrons in the atoms of the element with lowest energy are written first before those with higher energy levels. Thus, the electronic configuration shows the electrons in the atoms of the element arranged in order of increasing energies.

The electronic configuration of atoms are given as

A = 1s² 2s² 2p⁶ 3s¹

B = 1s² 2s² 2p⁶ 5s¹

The number of electrons in both the elements is 11. Therefore, their atomic number is also the same i.e, 11. So, both the elements are the same.

To know more about atomic number:

brainly.com/question/8645622

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4 0

2 years ago