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lapo4ka [179]
3 years ago
8

Which atom in a water molecule has the greatest electronegativity??

Physics
1 answer:
galina1969 [7]3 years ago
7 0
The oxygen atom in a water molecule has the greatest and or highest electronegativity, it has the greatest ability of all of the elements in a water molecule, 2, to pull electrons closer towards its own atom.
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A yellow ball with a mass of 2 kg is rolling across the floor at 3 m's A red ball with a mass of 3 kg is rolling across the
Wewaii [24]
Answer:
The red ball has a greater kinetic energy, because it has a greater mass

Explanation:
Mass is directly proportional to kinetic energy, as one increases, so does the other
6 0
3 years ago
Read 2 more answers
According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai
GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, \Delta v=3.5\times 10^3\ m/s

According to Heisenberg uncertainty principle,

\Delta x.\Delta p=\dfrac{h}{4\pi}

\Delta x is the uncertainty in the position of an electron

Since, \Delta p=m\Delta v

\Delta x=\dfrac{h}{4\pi.m \Delta v}

\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}

\Delta x=1.64\times 10^{-8}\ m

So, the uncertainty in its position is 1.64\times 10^{-8}\ m. Hence, this is the required solution.

6 0
3 years ago
A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
3 years ago
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0
3 years ago
What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original
DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

q₁ and q₂ are charges

r is the distance between charges

When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

F'=\dfrac{kq_1'q_2'}{r'^2}

Apply new values,

F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F

So, the new force becomes 64 times the initial force.

7 0
2 years ago
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