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3 years ago

Write the relation of m² with its a multiple

Physics

1 answer:

Harman [31]3 years ago

5 0

Explanation:

The square meter is the SI-derived unit of area. It has a symbol m² (33A1 in Unicode). It is defined as the area of a square whose sides measure exactly one metre.

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A particular star is white. Another star that is much hotter is expected to be ______.

jenyasd209 [6]

Blue

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4 years ago

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Applied force is the force of support exerted by an object that holds up another

beks73 [17]

Answer:

True, applied force is the force of support exerted by an object that holds up another object.

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4 years ago

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You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using e

Mashutka [201]

You are working for a manufacturing company, which is mathematically given as

$m=3\sqrt{2}$
$m=\frac{15\sqrt{5}}{16}$
x=0.747a
$m/n=\frac{(x^2+a^2)3/2}{x^3}$

<h3>What is the value of m that will place the movable bead in equilibrium at x-a a ....?</h3>

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

$K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3$

$m=3\sqrt{2}$

By force equilibrium

$K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}$

Therefore

$n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\$

$m=\frac{15\sqrt{5}}{16}$

$K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}$

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

By force equilibrium

$K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}$

$m/n=\frac{(x^2+a^2)3/2}{x^3}$

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3 0

2 years ago

4.) It was once recorded that a Jaguar left skid marks which were +300 m in length.

torisob [31]

Answer:

Initial velocity of the jaguar: 49 $\frac{m}{s^2}$ (answer d)

Explanation:

Considering that this uniformly accelerated problem (with negative acceleration since the jaguar was reducing its velocity to full stop), does not include the time the jaguar was skidding , we can use the kinematic equation that doesn't include time, but relates velocities (initial and final) with the acceleration ( $a$ ), and the distance "D" covered during the accelerated motion:

$v_f^2-v_i^2=2\,a\,D$

For our problem, the initial velocity ( $v_i$ is our unknown, the final velocity is zero ( $v_f = 0$ - since the jaguar stops in the process), the negative acceleration is given as $a=-4\,\frac{m}{s^2}$ , and the distance D of the skid marks is said to be 300 m in length. Therefore: