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3 years ago

10

Which values are equivalent to the fraction below ? check all that apply. 2^4/2^7

2 answers:

AnnyKZ [126]3 years ago

6 0

Answer:

2^-3, (1/2)^3, 1/8 apex

Explanation:

Irina18 [472]3 years ago

5 0

2^4/2^7 = 16/128 = 0.125
(1/2)^3= 0.125
1/8= 0.125
a and f are equivalent

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A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and

kvv77 [185]

Answer:

(a) 1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

$\overrightarrow{d_{1}}=580\widehat{j}$

$\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}$

$\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}$

The resultant displacement is given by

$\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}$

$\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}$

magnitude of the displacement

$d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m$

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

$tan\theta =\frac{1294.18}{35.36}=36.6$

θ = 88.44°

4 0

3 years ago

Based on its orbit, which planet behaves the least like the others?

Gekata [30.6K]

I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps

3 0

3 years ago

If V is the original volume, V' is the new volume, T is the original Kelvin temperature, and T' is the new Kelvin temperature, h

madam [21]

Answer:

$\frac{V}{T} = \frac{V'}{T'}$

Explanation:

Given the following data;

Original volume = V

New volume = V'

Original temperature = T

New temperature = T'

To write an expression for Charles's law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles law is given by the formula;

$\frac {V}{T} = K$

$\frac{V}{T} = \frac{V'}{T'}$

4 0

3 years ago

zhannawk [14.2K]

Average speed= total distance/total time =12km/h

6 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago