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3 years ago

What happens if you add additional,solid NaCl after the maximum has been reached?

Physics

2 answers:

Lina20 [59]3 years ago

7 0

Your answer would be he bond to the phosphate. Good luck

azamat3 years ago

3 0

it would bond to the phosphate

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What is the average power supplied by a 60.0 kg secretary running up a flight of stairs rising vertically 4.0 m in 4.2 s?

gayaneshka [121]

Answer:

9.8kW

Explanation:

Given data

Mass= 60kg

Hieght= 4m

Time= 4.2seconds

We know that the energy possessed is given as

PE=mgh

PE=60*9.81*4

PE= 2354.4 Joulse

Also, the expression for power is

Power=Energy*Time

Power= 2354.4*4.2

Power=9888.48 watt

Power= 9.8kW

4 0

3 years ago

Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2

lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

$F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C$

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 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:

 » Image distance :

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

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

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

 » Magnification :

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

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

» Multiply throughout by fv

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