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2 years ago

What happens when a candle burns?

Physics

1 answer:

slega [8]2 years ago

5 0

Answer:

oxygen is used up is the answer

Explanation:

These vaporized molecules are drawn up into the flame, where they react with oxygen from the air to create heat, light, water vapor (H2O) and carbon dioxide (CO2).

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Where are the magnetic fields strongest near a bar magnet?

KatRina [158]

The answer is near the poles.

3 0

2 years ago

Read 2 more answers

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 11.0 s. At th

mojhsa [17]

Answer:

Total angular displacement will be 19.998 radian

Explanation:

It is given that the washer starts from the rest and reach reach the speed of 2 rev/sec in 11 sec

So initial angular velocity $\omega _i=0rad/sec$

And final angular velocity $\omega _f=11rad/sec$

Time t = 11 sec

So angular acceleration $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{2-0}{11}=0.1818rad/sec^2$

So angular displacement in this 11 sec

$\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times 11+\frac{1}{2}\times 0.1818\times 11^2$

$\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times 11+\frac{1}{2}\times 0.1818\times 11^2=10.99radian$

Now the washer slows down and stops in 9 sec

So final angular velocity = 0 rad/sec

So angular acceleration $\alpha =\frac{0-2}{9}=-0.222rad/sec^2$

So angular displacement $\Theta =2\times 9-\frac{1}{2}\times 0.222\times 9^2=8.991radian$

So total displacement in 20 sec = $=10.999+8.999=19.998radian$

3 0

3 years ago

Can anyone please help me with this one

Vinvika [58]

Ur in k12 ALVA too???

6 0

3 years ago

A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.

arsen [322]

The solution for the acceleration of gravity is given as

$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

This is further explained below.

<h3>What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?</h3>

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$

$R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}$

In conclusion, acceleration due to gravity at this point will be

$g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

for $h_{2}=7900 \mathrm{~km}$

$R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

Read more about acceleration due to gravity

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5 0

2 years ago

What force was applied to an object if 35 J of work was done and the object moved 7m ?

den301095 [7]

The force applied to the object is 5 n

5 0

3 years ago