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hichkok12 [17]
3 years ago
14

Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?

Physics
1 answer:
Blababa [14]3 years ago
7 0

The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

When you look at an object located 1 light year away from you, you see it as it was 1 year ago.

If a star located 10 light years away from us suddenly brightens, or dims, or explodes, we see the event <em>10 years later.</em>

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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v
Aleks [24]

Answer:

0.116 m^3/s

Explanation:

The volume flow rate of a fluid in a pipe is given by:

Q=Av

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2

So, the volume flow rate is

Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s

8 0
4 years ago
An organism that obtains organic food molecules by eating other organisms or their by-products
nikitadnepr [17]

Answer:

heterotrophs

Explanation:

According to the parameters established by biology, all living beings that require others to feed themselves are considered heterotrophs, that is, they are not able to produce their food within their organism but rather they must consume elements of nature already constituted as food, already synthesized by other organisms. Among the most prominent heterotrophs, all animals, bacteria and humans stand out.

The term heterotroph comes from the Greek, language in which the prefix hetero means different and trophies means food. In this way, the heterotroph is one that feeds on elements other than one, which takes elements from nature, from the surrounding space to feed. While autotrophic beings have the ability to synthesize inorganic elements such as light, water, carbon dioxide and convert them into food; Heterotrophic beings do not have that capacity, so they must consume plants (in the case that they are herbivores) or animals that have already consumed those plants (that is, in the case that they are carnivorous). In other words, animals and humans always need to feed on other living beings, they could never do so only from inorganic elements such as water.

4 0
3 years ago
If each pull-up requires 300 J and Ben does a pull-up in 2 seconds, what is his power? 150 watts 300 watts 600 watts 750 watts
leonid [27]

Answer:

150 watts

Explanation:

300/2 = 150 watts

6 0
2 years ago
Read 2 more answers
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
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