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Molodets [167]
2 years ago
6

Hey can anyone please help me with this it’s due in few hours and I’m stuck with ittt

Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: 800* 24 = 1,92 *10^4kg

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's 1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N

2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

3. Yet again, \rho gh. 1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa

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The correct answer to the question is : 9375 N.

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}

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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

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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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