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4 years ago

_____ is a chemical property of matter.

Physics

1 answer:

german4 years ago

8 0

Answer:

it is chemical reactivity (D)

Explanation:

it is because chemical properties are properties that can be measured and observed only when matter undergoes a change to become an entirely different kind of matter.Which they include Chemical reactivity,flammable,and the ability to rust.And Chemical reactivity is the ability of matter to react chemically with other substances.

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Express 0.00000000062 kg in scientific notation. 6.2 x 1010kg 6.2 x 10-10kg 0.62 x 10-9kg

PilotLPTM [1.2K]

I believe the correct answer from the choices listed above is the second option. The scientific notation of the measurement 0.00000000062 kg would be <span>6.2 x 10^-10 kg. Scientific notation is used to express too large and too small values of numbers. Hope this helps. Have a nice day.</span>

7 0

3 years ago

Read 2 more answers

What two units of measure are necessary for describing speed

Shtirlitz [24]

Distance and time, distance because that's how far and time because that's how long

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4 years ago

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GenaCL600 [577]

Answer:

Explanation:

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3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

3 years ago