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Mariulka [41]
3 years ago
15

El Niño occurs every 2 to 12 years. true false

Physics
2 answers:
wlad13 [49]3 years ago
8 0

yep this is true i took the test

Novosadov [1.4K]3 years ago
5 0
True
It is True I took the test
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Which of the following would be used in luminosity calculations?
Daniel [21]

Answer:

1) joule

2) kgm^{2}/s^{2}

3) 10\%

Explanation:

1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second (\frac{J}{s}), which is equal to Watts (W).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work W is expressed as force  F multiplied by the distane  d :

W=F.d

Where force has units of  kgm/s^{2} and distance units of m.

If we input the units we will have:

W=(kgm/s^{2})(m)

W=kgm^{2}/s^{2}  This is 1Joule (1 J) in the SI system, which is also equal to 1 Nm

3) The formula to calculate the percent error is:

\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%

Where:

V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2} is the experimental value

V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2} is the accepted value

\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%

\% error=10.04\% \approx 10\% This is the percent error

8 0
3 years ago
I need help plz and thank you❤
anzhelika [568]
I think:
In motion- 40
Not moving- 20
6 0
3 years ago
Please help, thanks!
statuscvo [17]
Because it demonstrates the relationship between a body and the forces acting upon it, and its motion in response to those forces. [Hope that helps]
7 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy.  That idea is easy to apply to this question.

8 0
4 years ago
Read 2 more answers
9.58 A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. T
andrey2020 [161]

Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

4 0
3 years ago
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