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3 years ago

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A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli

gible friction down the hill, through a circular loop of radius 20 m (Positions B,C,and D) and along a horizontal track (to position E). Describe the scientific principles involved in this scenario.

1 answer:

Andrews [41]3 years ago

4 0

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

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The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

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Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

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Answer:

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Required

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$A \to rad/s$

$Bt^2 \to rad/s$

The unit of t is (s); So, the expression becomes

$B * s^2 \to rad/s$

Divide both sides by $s^2$

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