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3 years ago

5

The illustration shows a roller coaster and indicates four different positions the car might be at as it moves along the track.

Identify at which point in the roller coaster's journey does it have the least potential energy and explain why. w

1 answer:

DerKrebs [107]3 years ago

4 0

Answer:

I got this question on Ap3x. The answer is Car C...I got it correct

Explanation:

This is because Car C is at the lowest point with the lowest amount of potential energy. Potential energy is stored at it's highest when it has the "potential" to fall, move, or etc. Car C seems to have gone farther down from the high point of the slope, meaning that most of the potential energy transformed into kinetic energy. All in all, Car C has the least potential energy. (Please give Brainliest, or not...your choice but this is the first question that I answered)

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

What happens when sound waves move faster?

Lady bird [3.3K]

Hnjjjjj hejehtjriwje a wkirrhnemekwhr. Rthejb wine rnrie

6 0

3 years ago

Hatshy [7]

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

5 0

4 years ago

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M $_S$ = 1.99 × 10³⁰ kg

Mass of the neutron star

M $_N$ = 2( M $_S$ )

M $_N$ = 2( 1.99 × 10³⁰ kg )

M $_N$ = ( 3.98 × 10³⁰ kg )

Radius of neutron star R $_N$ = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω $_N$ .

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM $_N$ = / R $_N$ ² = mR $_N$ ω $_N$ ²

ω $_N$ ² = GM $_N$ = / R $_N$ ³

ω $_N$ = √(GM $_N$ = / R $_N$ ³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω $_N$ = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω $_N$ = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω $_N$ = √ 120831133.3636777

ω $_N$ = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0

3 years ago

Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.

Aloiza [94]

Answer and Explanation:

Comparison between the Titan's atmosphere and earth atmosphere

Titan atmosphere is more denser than the earth atmosphere
The quantity of nitrogen is more in titan atmosphere than earth atmosphere titan atmosphere have about 98 % of nitrogen in the other hand earth atmosphere has only 78 %
There are no oxygen present in titan atmosphere while in earth atmosphere it is present
Its air is not suitable for breath but in earth atmosphere we can breath

5 0

3 years ago