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marusya05 [52]
3 years ago
9

What is the following product √10 * √10 10 10√10 100 2√10

Physics
2 answers:
insens350 [35]3 years ago
8 0
√10(√10) = √10^2

The root and the square cancels:
√10^2 = 10

10 is your answer

hope this helps
qwelly [4]3 years ago
7 0

Answer:

The result of the product √10*√10 is 10

Explanation:

To solve this product, it is first appropriate to apply the property of root multiplication with the same index.  This property indicates that multiplying two roots with the same index is the same as multiplying on a single root with that index. In this case this property is applied as follows:

\sqrt{10} *\sqrt{10} =\sqrt{10*10}

Now you can think of the next step in two ways. One of them is to multiply within the root and then solve the root as follow:

\sqrt{10*10} =\sqrt{100} =10

Another way is to express the values ​​within the root as power, because they are the same number.

\sqrt{10*10} =\sqrt{10^{2} }

The root is the opposite operation to the power, then the root nullification property can be applied: if you have an elevated number or variable to an exponent that is within a root with the same index, the power with the root is voids:

\sqrt{10^{2} } =10

Note that in both cases the result is the same, so the method you choose to resolve this account is indistinct. <em><u>The result of the product √10*√10 is 10</u></em>.

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3 years ago
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
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Answer:

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91.64 km high material would go on earth if it were ejected with the same speed as on Io.

Explanation:

According to Newton Law of gravitation:

g=\frac{Gm}{r^2}

Where:

G is gravitational constant=6.67*10^{-11} m^3/kg.s^2

For Moon lo g is:

g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2

According to law of conservation of energy

Initial Energy=Final Energy

K.E_i+mgh_i=K.E_f+mgh_f

\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}

For Jupiter's moon Io:

Velocity is given by:

v_0_M=\sqrt{2g_Mh_f_M}

For Earth Velocity is given by:

v_0_E=\sqrt{2g_Eh_f_E}

Now:

v_o_M=v_o_E

\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}

g_E=9.8 m/s^2

g_m=1.7962 m/s^2, As\ Calculated\ above

h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

8 0
3 years ago
Chemistry Sem 2
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Answer:

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777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

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If distance 2, r' = 25 m

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I=\dfrac{P}{4\pi r^2}.........(1)

Let I' is the intensity at r'. So,

I'=\dfrac{P}{4\pi r'^2}............(2)

From equation (1) and (2) :

I'=\dfrac{Ir}{r'^2}

I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}

I'=1.98\times 10^{-9}\ W/m^2

Intensity level is given by :

dB=10\ log(\dfrac{I'}{I_o}), I_o=10^{-12}\ W/m^2

dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})

dB = 32.96 dB

Hence, this is the required solution.

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3 years ago
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