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solong [7]
3 years ago
11

The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.

Physics
1 answer:
wolverine [178]3 years ago
5 0

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

\rm 1 \: g =  {10}^{ - 3}  \: kg \\  \\  \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6  \times  {10}^{ - 3} \ kg \\  \\  \rm = 1.440 \times 10^{6 - 3} \ kg \\  \\  \rm = 1.440 \times 10^3 \ kg \\  \\ \rm = 1.440 \times 1000 \ kg \\  \\ \rm = 1440 \ kg

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an object tied to the end of the string moves in a circle . the force exerted by the string depends on the mass of the object it
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3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
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