Ask question

Ask question

All categories

notsponge [240]

2 years ago

14

What is the most significant change in the comet's energy as it moves from

1 answer:

svetoff [14.1K]2 years ago

3 0

The most significant change in the comet's energy as it moves from point D to point A KE increases, GPE decreases.Option C is correct.

<h3>What are comets?</h3>

Comets are frozen leftovers from the formation of the solar system composed of dust, rock, and ice.

They range from a few miles to tens of miles wide, but as they orbit closer to the Sun, they heat up and spew gases and dust into a glowing head that can be larger than a plane.

As the comet transitions from point D to point A, the biggest energy shift is seen. GPE declines as KE rises. The right answer is C.

Hence option C is correct.

To learn more about the comet refer;

brainly.com/question/12443607

#SPJ1

You might be interested in

barxatty [35]

Answer:

NE DIYON INGILIZ MISIN SEN

7 0

3 years ago

What do magnets never do?<br><br><br> This is science btw.

Anika [276]

Answer:

they..don't cook foood

8 0

2 years ago

A 65 Kg skier starts at rest at the top of a 150 m long hill that has an incline of 28 degrees. How fast will she be going at th

yan [13]

Answer:

34 m/s

Explanation:

Potential energy at top = kinetic energy at bottom + work done by friction

PE = KE + W

mgh = ½ mv² + Fd

mg (d sin θ) = ½ mv² + Fd

Solving for v:

½ mv² = mg (d sin θ) − Fd

mv² = 2mg (d sin θ) − 2Fd

v² = 2g (d sin θ) − 2Fd/m

v = √(2g (d sin θ) − 2Fd/m)

Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:

v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))

v = 33.9 m/s

Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.

8 0

3 years ago

How long will it take a plane to fly 1256km<br> if it travels 500km/hr?

expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

d is the distance

v is the velocity

From the question we have

$t = \frac{1256}{500} \\ = 2.512$

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0

2 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

3 years ago