All categories

4 years ago

What happens if you add additional,solid NaCl after the maximum has been reached?

Physics

2 answers:

Lina20 [59]4 years ago

7 0

Your answer would be he bond to the phosphate. Good luck

azamat4 years ago

3 0

<span>it would bond to the phosphate

</span>

You might be interested in

Electromagnets are created by

Scrat [10]

I electromagnets, the magnetic field is turned off when there is no induced current. So, electromagnets act as magnets only when current is induced in an insulated wire wrapped around a material mage of ferromagnetic material such as iron core rotating in a magnetic field.

So, option D is the correct one.

If you wanna know more about electromagnets, write down in comments {:

6 0

3 years ago

Read 2 more answers

Someone help me please

Maru [420]

H I and u your weclome

8 0

3 years ago

As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly and at -4.1m/s^2 as it slows from 9.0

Tju [1.3M]

Hope this helps you.

5 0

3 years ago

Ivenika [448]

The ant's resultant velocity is 0.51 m/s at 78.7°

Since the ant is moving with a velocity of 0.1 m/s in the +x direction relative to the treadmill, and the treadmill is moving 0.5 m/s in the +y direction, the magnitude of the resultant velocity of the ant relative to the ground (since both directions are perpendicular) is thus

V = √(x² + y²)

= √[(0.1 m/s)² + (0.5 m/s)²]

= √[0.01 m²/s² + 0.25 m²/s²]

= √(0.26 m²/s²)

= 0.51 m/s.

The direction is Ф = tan⁻¹(y/x)

= tan⁻¹(0.5 m/s ÷ 0.1 m/s)

= tan⁻¹(5)

= 78.69°

≅ 78.7°

The ant's resultant velocity is 0.51 m/s at 78.7°

Learn more about resultant velocity here:

brainly.com/question/15093278

7 0

3 years ago

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

3 years ago