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3 years ago

10

A carnival ride has a 2.0m radius and rotates once each .90s. Find the centripetal acceleration

1 answer:

Paraphin [41]3 years ago

3 0

We know that centripetal acceleration is nothing but the ratio of the square of the tangential velocity to that of the radius vector.
a=v*v/r=ωωrr/r
=ωωr
=2πf2πfr
=2π2πr/TT
=97m/secsec

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son4ous [18]

Answer:

It has been converted into thermal energy due to friction

Explanation:

According to the law of conservation of energy, energy cannot be created nor destroyed, but only transformed from one form into another.

Applied to this problem, it means that the total initial energy of the spring-toy system must be conserved.

Therefore:

- At the beginning, the total energy stored in the spring is 10 J

- After the toy is released, the total energy must still be 10 J.

In reality, we are told that the kinetic energy of the car is only 8 J. The other 2 J have not been destroyed, but they have been converted into thermal energy, due to the presence of frictional forces that act against the motion of the toy car.

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3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

Daniel [21]

Answer: $81.57\mu V$

Explanation:

Given

radius of circular region r=1.50 mm

$A=\pi r^2=7.069\times 10^{-6}\ m^2$

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time t=130 ms

Flux is given by

$\phi =B\cdot A$

change in Flux $d\phi =(B_f-B_i)A$

Emf induced is $e=\frac{\mathrm{d} \phi}{\mathrm{d} t}$

$e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}$

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4 years ago

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

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Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

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Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

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3 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

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In this problem, the work done by the system is

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with a negative sign as well because it is released by the system.

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olasank [31]

<em>Answer: </em>tellurium (Te)

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Fourth energy level = 18

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3 years ago

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