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2 years ago

If a substance was described as smooth, square, and tan, what properties are

Physics

2 answers:

Nastasia [14]2 years ago

8 0

D.) is the correct answer!

Sergeeva-Olga [200]2 years ago

4 0

D. Physical properties

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How to find angular velocity from linear velocity and radius.

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Answer:

ω=v/r.

Explanation:

<em><u>angular velocity= linear velocity/radius</u></em>

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2 years ago

The emt must assume that any unwitnessed water-related incident is accompanied by:________

natta225 [31]

The EMT must assume that any unwitnessed water-related incident is accompanied by potential spinal damage.

<h3>What is spinal damage?</h3>

Nerves or the spinal cord in any way damaged at the end of the spinal canal.

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Below the damage site, a spinal cord injury frequently results in a lifelong loss of strength, feeling, and function.

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Symptom-reducing medications and spinal stabilisation surgery are used as treatments.

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1 year ago

An object is pulled with two forces, 10 N northward and 15 N southward. The direction of the net force is to the An object is pu

ValentinkaMS [17]

Answer:

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Explanation:

For any question related to newons law of motion first draw the free body diagram(FBD),

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3 years ago

Why echos are undesirable in a big hall

KatRina [158]

Answer:

Helloooo

Sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation. In an auditorium or big hall excessive reverberation is highly undesirable

Explanation:

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3 years ago

Read 2 more answers

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago