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Phantasy [73]
2 years ago
3

Does taking 1 mL of solution Y (pure bleach) out of the first beaker change the solution’s concentration? What would the concent

ration of the second beaker be? If Beaker #2 is a 1% solution of Y, and Beaker #3 is 1/100th of that in the final solution, what percentage solution would Beaker #3 be?
Physics
2 answers:
chubhunter [2.5K]2 years ago
7 0
-No
-1/100th of solution Y
-0.01% bleach
Cloud [144]2 years ago
7 0

-No

-1/100th of solution Y

-0.01% bleach


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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens woul
Brilliant_brown [7]

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

7 0
2 years ago
A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he
Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

7 0
2 years ago
A man of mass 80kg stands next to a stationary ball of mass 4kg on a frictionless surface. He kicks the ball forward along the s
scZoUnD [109]
Momentum is conserved.

p = m₁v₁ + m₂v₂ = constant

Momentum before the collision(kick):

p = 80 * 0 + 4 * 0 = 0

Momentum after the collision:
p = 80 * v₁ + 4 * 15 = 0

Solve for v₁.

3 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
2 years ago
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