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3 years ago

. Friction is a rubbing force that ___________ a spinning yo-yo.

Physics

2 answers:

katovenus [111]3 years ago

5 0

Answer:

i think add energy

Explanation:

Advocard [28]3 years ago

3 0

The yo-yo speeds up when you rub it

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if the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity?

erastova [34]

The answer to this question should be: The accuracy in measuring its velocity decreases

Hope I helped

7 0

3 years ago

A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?

eduard

First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock, then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

√(1 - v²/c²) .

You said that v/c = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² = 1 - 0.7225 = 0.2775

√(1 - v²/c²) = √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 = <em>18.983 years </em>

I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

8 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

The New angular speed is $w_f = 2.034 rad/s$

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago

in 4 seconds, a cheetah can run 112 meters. How fast does this cheetah have to be running? A: 448 m/s || B: 28 m/s || C: 56 m/s

Vladimir [108]

Answer:

B: 28m/s

Explanation:

Use the speed formula

Speed =distance / time

8 0

3 years ago

Read 2 more answers

A test charge of -1.4 × 10-7 coulombs experiences a force of 5.4 × 10-1 newtons. Calculate the magnitude of the electric field c

mr_godi [17]

I would say C 5.4 x 10-1 Newtons/coulomb

3 0

3 years ago

Read 2 more answers