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3 years ago

What is the only bone of the skull that moves?

Physics

2 answers:

NikAS [45]3 years ago

8 0

Answer:

mandible

Explanation:

which is also the lower jaw and it allows the mouth to open and close

hope i helped

Darina [25.2K]3 years ago

3 0

Answer:

Mandible.

Explanation:

The mandible in the bone that allows you to open and close your mouth, otherwise known ad the jawbone.

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

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$t'=\frac{342760}{1362}$

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3 years ago

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

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2 meters per second²

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2 years ago

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The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm= $8\times 10^{-2} m$

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Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid= $8\times 10^{-2} m$

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$B=\frac{\mu_0NI}{l}$

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$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

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2 years ago