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3 years ago

6

A helpful hand please

1 answer:

Tpy6a [65]3 years ago

3 0

$\bf s=\sqrt{2gh}\quad 
\begin{cases}
s=speed\\
g=gravity\\
h=height\\
-----\\
s=34\\
g=32
\end{cases}\implies s^2=2gh\\\\\\ \cfrac{s^2}{2g}=h\implies \cfrac{34^2}{2\cdot 32}=h$

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What is S7 for 6 – 24 + 96 – 384 + ... ?

Dmitry [639]

6 - 34 + 96 - 384 +...
a 1 = 6 , a 2 = a 1 * ( - 4 ) , a 3 = a 2 * ( - 4 ), a 4 = a 3 * ( - 4)...
a 5 = - 384 * ( - 4 ) = 1,536
a 6 = 1536 * ( - 4 ) = - 6,144
a 7 = - 6,144 * ( - 4 ) = 24,576
This is a sum of a geometric sequence:
S 7 = 6 - 24 + 96 - 384 + 1,536 - 6,144 + 24,576 = 19,662
Another way to solve this:
S 7 = a 1 * ( r^n - 1) / ( r - 1 ) = 6 * (( - 4) ^7 - 1 )/ (- 4 - 1)=
= 6 * (-16,385) / (-5 ) = 19,662

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2 years ago

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The focal lengths of the objective lens and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separat

VladimirAG [237]

Answer:

- 103.7

Step-by-step explanation:

Given:

Focal length of the eyepiece, f = 2.0 cm

Focal length of the objective lens, f' = 0.50 cm

Separation for minimum eyestrain = 6,0 cm

Image distance, v = - 25 cm

Now, from the lens formula,

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here, u is the object distance

on substituting the respective values, we get

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or

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also, the separation is adjusted for minimum eyestrain,

therefore, image distance for the objective lens, v' = 6 - 1.852 = 4.148 cm

Now, for the objective lens

using the lens formula, we get

$\frac{1}{0.5}=\frac{1}{u'}+\frac{1}{4.418}$

Here, u' is the distance between the physical object and objective lens

or

u' = 0.568 cm

Thus,

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or

m = $-\frac{25\times4.418}{0.5\times2}$

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2 years ago

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Answer:

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Step-by-step explanation:

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