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goblinko [34]
3 years ago
15

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma

gnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices
Physics
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  a_t = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\

where;

a_c is the radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2

The radial acceleration relates to speed of particle in the following equations;

a_c = \frac{v^2}{r}

where;

v is the speed of the particle

v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s

Therefore, the speed of the particle is 2.86 m/s

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You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100
Vikki [24]

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a = F/m

a = 250/85

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using the equation

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consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

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