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3 years ago

In a mixture, the ingredients intermingle and

Physics

1 answer:

nirvana33 [79]3 years ago

7 0

They do not react with or chemically bond to each other.

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A distressed car is rolling backward, downhill at 3.0 m/s when its driver finally manages to

konstantin123 [22]

Answer:

Explanation:

Acceleration is equal to the change in velocity over the change in time, or

$a=\frac{v_f-v_i}{t}$ where the change in velocity is final velocity minus initial velocity. Filling in:

$3.0=\frac{v_f-(-3.0)}{6.0}$ Note that I made the backward velocity negative so the forward velocity in our answer will be positive.

Simplifying that gives us:

$3.0=\frac{v_f+3.0}{6.0}$ and then isolating the final velocity, our unknown:

3.0(6.0) = v + 3.0 and

3.0(6.0) - 3.0 = v and

18 - 3.0 = v so

15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.

5 0

3 years ago

HELP PLZZZZZZ

ValentinkaMS [17]

Hi there!

We know that:

U (Potential energy) = mgh

We are given the potential energy, so we can rearrange to solve for h (height):

U/mg = h

g = 9.81 m/s²

m = 30 g ⇒ 0.03 kg

0.062/(0.03 · 9.81) = 0.211 m

8 0

3 years ago

A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then

Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J