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2 years ago

The wave shown below is produced in a rope.

Physics

2 answers:

Kay [80]2 years ago

6 0

Answer:

<h3>B) A dot vector drawn up and vector drawn down.</h3>

Explanation:

Brainliest?

olga_2 [115]2 years ago

5 0

Answer:

A dot vector drawn up and vector drawn down.

Explanation:

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The wavelengths for visible light rays correspond to which of these options?

ryzh [129]

Answer:

sorry about the other person but its b

Explanation:

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2 years ago

The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter

ryzh [129]

Answer:

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2 years ago

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The main illustration in the video shows the life track of a one-solar mass star. Each point along this track represents _______

REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

<h3>What is the one-solar mass star?</h3>

A star having a mass equal to the mass of the Sun is called a one-solar mass star.

Its life track shows the luminous intensity as well as the surface temperature.

Learn more about one-solar mass star.

brainly.com/question/14984575

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1 year ago

A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu

bija089 [108]

Answer:

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed $v_1$ and changes it to $v_2$ . The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

$p=mv$

The initial and final momentums are, respectively

$p_1=mv_1,\ p_2=mv_2$

The change of momentum is

$\Delta p=p_2-p_1=m(v_2-v_1)$

It is numerically equal to the Impulse J

$J=\Delta p$

$J=m(v_2-v_1)$

We are given

$m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s$

The impulse the car experiences during that time is

$J=2498(2.6-17.1)=2498(-14.5)$

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

8 0

3 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

2 years ago