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3 years ago

14

Which has a greater effect on our tides, the sun or the moon?

2 answers:

charle [14.2K]3 years ago

7 0

The moon has a greater effect, because it has a gravitational pull also known as gravity.

charle [14.2K]3 years ago

6 0

I’m pretty sure the moon

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Sonrisa owns a 300 W television. If the total energy usage for February is 32.4 kWh, how many hours per week does Sonrisa watch

Zielflug [23.3K]

the correct answer is 27 hours per week :) hope this helps

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4 years ago

Read 2 more answers

An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?

andrey2020 [161]

$From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.$

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3 years ago

What formula is used to find an objects acceleration

ololo11 [35]

Acceleration = velocity / time.

7 0

3 years ago

Two double bonds means that the total number of electrons being shared in the molecule is

Tema [17]

Answer:

9 terms. In carbon dioxide (CO2), there are two oxygen atoms for each carbon atom. Each oxygen atom forms a double bond with carbon, so the molecule is formed by two double bonds. Two double bonds means that the total number of electrons being shared in the molecule is.

Explanation:

3 0

3 years ago

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago