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Vladimir [108]
2 years ago
12

6 latter word and it has a s and a I and it has mass (9.______________liquids, and gases all have mass.)

Physics
1 answer:
Firdavs [7]2 years ago
7 0

Answer:

Solids

Explanation:

Solids, liquids, and gases all have mass.

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A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
Help ASAP with equation and answer pls. Numbers 2 3 4
scoray [572]

Answer:

2. 200N

3.50kg

4.700N

Explanation:

Weight is another word for the force of gravity

Weight is a force that acts at all times on all objects near Earth.

F=m*g

where g=acceleration due to gravity

2. due to the gravitational fields of the earth , assume gravitational acceleration=10m/s2

F=20*10= 200N

3.same as above

mass=Force/gravitational acceleration

mass=500/10 = 50kg

4.force=mass*gravitational acceleration

force=70*10=700N

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3 years ago
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fgiga [73]

Answer:

B is the answer

Explanation:

0-/-<

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3 years ago
How much time would it take for the sound of thunder to travel 2000 meters if sound travels of 330 meters per sec
Lubov Fominskaja [6]
2000÷330=6.06 repatant so the answer would be about 6.06 seconds
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3 years ago
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Answer:

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