All categories

3 years ago

Is pure water a acid or a base?

Physics

2 answers:

Xelga [282]3 years ago

5 0

Answer:

NEITHER.

Explanation:

it is neutral

Flura [38]3 years ago

3 0

Answer:

The pure water is neutral since its PH is 7

You might be interested in

laila [671]

Answer:

Explanation:

A dreams

.......wkkwkwkwkwkwnwnsksk

4 0

3 years ago

Which statement is true about the formation of bounds

MariettaO [177]

The formation of bonds releases energy.

4 0

3 years ago

What causes air to become less dense and rise?

aalyn [17]

Answer:

That force is 'gravity'.

8 0

4 years ago

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0Â° above the horizontal. it strike

Anni [7]

<span>x = 129.9 m y = 30.9 m First, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m</span>

8 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.

d. The speed of the particle at $t=7.00\,\mathrm s$ is the magnitude of the velocity at this time:

$\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath$

$\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}$

Then its speed at this time is

$\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}$

4 0

3 years ago