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3 years ago

6

a ladybug sits at the outer edge of a merry-go-round and a gentleman bug sits halfway between her and the axis of rotation. The

merry-go-round makes a complete revolution once each second. What is the gentleman bug's angular speed?

1 answer:

forsale [732]3 years ago

6 0

The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)

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The toy car. An object that isn't moving has no momentum

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2 years ago

Pls answer I will make brainlist

Semenov [28]

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3 years ago

A cat runs in a straight line. Which of the following statements about the cat's motion must be true?

alexira [117]

Answer:

I would think it's A cat runs in a straight line.

Explanation:

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3 years ago

A record spins at 33 rpm (revolutions per minute), which is an angular velocity of about 3.46 radians per second. What is the ap

lubasha [3.4K]

Hi there!

We can use the following equation to relate angular velocity to linear velocity.

$v = \omega r$

v = linear velocity (m/s)

ω = angular velocity (3.46 rad/sec)

r = distance from axis of rotation (.12 m)

Plug in the given values.

$v = (3.46)(.12) = \boxed{.415 \frac{m}{s}}$

6 0

2 years ago

In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with t

Mamont248 [21]

Answer:

a) $h_{max}=14536.16 m$

b) $h = 15687.9 m$

c) $PD=7.62\%$ The estimate is low.

Explanation:

a) Using the energy conservation we have:

$E_{initial}=E_{final}$

we have kinetic energy intially and gravitational potential energy at the maximum height.

$\frac{1}{2}mv^{2}=mgh_{max}$

$h_{max}=\frac{v^{2}}{2g}$

$h_{max}=\frac{43^{2}}{2*0.0636}$

$h_{max}=14536.16 m$

b) We can use the equation of the gravitational force

$F=G\frac{mM}{R^{2}}$ (1)

We have that:

$F = ma$ (2)

at the surface G will be:

$G=\frac{gR^{2}}{M}$

Now the equation of an object at a distance x from the surface.

is:

$F=\frac{mgR^{2}}{(R+x)^{2}}$

$m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}$

Using that dv/dt is vdx/dt and integrating in both sides we have:

$v_{0}=\sqrt{\frac{2gRh}{R+h}}$

$h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}$

$h=15687.9$

c) The difference is:

So the percent difference will be:

$PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%$

$PD=7.62\%$

The estimate is low.

I hope it helps you!

7 0

3 years ago