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1 year ago

Match the concepts in Column 1 to the definitions and explanations in Column 2.

Physics

1 answer:

Akimi4 [234]1 year ago

3 0

Answer:

SRY I CAN'T ANS :( and hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

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What is the momentum of a 546,540 kg train that is travelling at 7.8 m/s

lara [203]

p=mv so wouldn't u multiply them?

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2 years ago

4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

Kazeer [188]

Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

and direction $\theta_R$ such that

$\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ$

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2 years ago

A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr

Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

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1 year ago

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon

Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle $\theta =51^{\circ}$

launch velocity $u=18 m/s$

center of mass continue to travel its original Path so it center of mass will be at a distance of

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{18^2\sin 102}{9.8}$

$R=32.33 m$

Center of mass will be at $x=32.33 m$

(b)if one of the piece will be at x=26 m then other will be at

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}$

$32.33=\frac{26+x_0}{2}$

$x_0=38.66 m$

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2 years ago

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Flura [38]

A, it's better to take a break to let your body rejuvenate.

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