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krok68 [10]
2 years ago
8

A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.

For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Throwing a discus. Part A For a point on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 ss of acceleration
Physics
1 answer:
charle [14.2K]2 years ago
5 0

Answer: 0.00024\ m/s^2

Explanation:

Given

Radius of flywheel is r=0.4\ mm

Angular acceleration \alpha=0.6\ rad/s^2

For no change in radius, tangential acceleration is  given as

\Rightarrow a_t=a\lpha \times r

Insert the values

\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

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3 years ago
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UNO [17]
V = IR

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3 years ago
You buy a 75 watt light bulb. the label means that
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4 0
3 years ago
A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a
kirill [66]

Answer:

θ = 13.16 °

Explanation:

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The friction force = (m g)/5

Now from work power energy

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work done by all forces = change in kinetic energy

( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²

(2  g sinθ - ( 2 g)/5 ) d = v² -  u²

take g = 10 m/s²

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5 0
2 years ago
If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:
madam [21]

Answer:

kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

Explanation:

Given the data in the question;

we know that;

Kinetic energy = 1/2.mv²

given that mass of the object is doubled; m1 = 2m

speed is halved; v1 = V/2

Now, New kinetic energy will be; 1/2.m1v1²

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Kinetic Energy = 1/2 [ 1/2mv² ]

Kinetic Energy = 1/2 [ KE ]

Therefore; kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

3 0
2 years ago
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