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3 years ago

9

What is the difference in sodium and chlorine

2 answers:

shtirl [24]3 years ago

8 0

sodium is a ionic compound

Chlorine is a nonmetal

(adding on to first answer)

SSSSS [86.1K]3 years ago

6 0

sodium is a metal in group 1

chlorine is a non metal in group 7

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Q10. Refer to the Condon table to answer question

Mnenie [13.5K]

Answer:

so you have a question

Explanation:

either way, have a nice day

3 0

3 years ago

You have finger tips but you dont have toe tips but you can tip toe explain pls

Arada [10]

Answer:

ummmmmmmm

Explanation:

this might be some type of trick question bro lol

or someone tryna play with you if the teacher gave you this question bro she deserves to be fired

4 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)

Lunna [17]

Answer:

a

The number of radians turned by the wheel in 2s is $\theta= 8\ radians$

b

The angular acceleration is $\alpha =14 rad/s^2$

Explanation:

The angular velocity is given as

$w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

This is mathematically evaluated as

$\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

$= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

$= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

$= 4 +4$

$\theta= 8\ radians$

Now generally the derivative of angular velocity gives angular acceleration

So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

This is mathematically evaluated as

$\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

$\alpha (2) = 2 +3(2)^2$

$\alpha =14 rad/s^2$

7 0

3 years ago

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time of computer A is $CT_{A} =200 ps$

Clock time of computer B is $CT_{B} =250 ps$

Effective CPI of computer A is $CPI_{A} =1.5$

Effective CPI of computer B is $CPI_{B} =1.7$

CPU time of A is

$CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec$

CPU time of B is

$CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec$

Hence Computer A is Faster by $\frac{425}{300} =1.41$

Computer A is 1.41 times faster than the Computer B

4 0

3 years ago