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schepotkina [342]

3 years ago

15

Which of the following is not an example of osmosis? Multiple Choice

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

Sugar in coffee dissolves.

Explanation:

In this case we have a solution, which is defined as separating what was bound in some way, by homogeneously mixing the molecules of a substance into a liquid. Therefore, it is the resulting homogeneous mixture after dissolving any substance in a liquid.

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

2 years ago

A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g

nata0808 [166]

Answer:

b

Explanation:

5 0

2 years ago

Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your f

Paul [167]

Answer:

Explanation:

lesgse in no

3 0

2 years ago

A proton moves at a constant velocity of 50m/s along the x axis ,through crossed electric and magnetic fields .The magnetic fiel

shusha [124]

Answer:

42 I THE CORRECT ANSWER

Explanation:

BECAUSE I KNOW

6 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

2 years ago