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2 years ago

Magnesium hydroxide is a common ___?

Physics

2 answers:

omeli [17]2 years ago

6 0

Answer:

Magnesium hydroxide is the inorganic compound with the chemical formula Mg(OH)2. Magnesium hydroxide is a common component of antacids, such as milk of magnesia, as well as laxatives

Explanation:

pls mark brainliest

diamong [38]2 years ago

4 0

Answer:

Hello There!!

Explanation:

The answer is=> It is a common component of antacids.

hope this helps,have a great day!!

~Pinky~

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A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has

Masja [62]

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$

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3 years ago

A device that uses electricity and magnetism to create motion is called a _________motor,magnet,generator . In a reverse process

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A device that uses electricity and magnetism to create motion is called a "Motor" (which converts electric energy into mechanical energy) & <span>In a reverse process, a device that uses motion and magnetism can be used to create "Electromagnetism".

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2nd Blank = Electromagnetism

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2 years ago

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Which planet will take the shortest revolution around the sun?

LUCKY_DIMON [66]

The planet closest to the sun; Mercury.

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3 years ago

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$