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Stells [14]
3 years ago
11

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

s 1000 kg. At this concentration, what mass of uranium is present in 1.2 mg of the earth's crust?
Physics
1 answer:
timama [110]3 years ago
7 0

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

For a sample of mass 1.2 mg, the amount of uranium is:

1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg

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A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force
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CALCULATION:

As per the question, the mass of the car  m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = \frac{D}{2}

                                                                  = \frac{200}{2}\ m

                                                                  = 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

                              Force F = \frac{mv^2}{R}

                                            =  \frac{1500\times (25)^2}{100}\ N

                                            = 9375 N.           [ANS}        

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.


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