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Citrus2011 [14]
3 years ago
8

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
<span>it would bond to the phosphate 

</span>
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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:

<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
What type of heat does not require matter?
Lana71 [14]
It would be Thermal Radiation
4 0
3 years ago
Read 2 more answers
In order to be considered a vector, a quantity must include what two components? A. Magnitude
shepuryov [24]

In order to be considered a vector, a quantity must include Magnitude (A) and Direction (D).

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3 years ago
Read 2 more answers
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