All categories

3 years ago

When two hydrogen atoms bond, the positive nucleus of one atom attracts the

1 answer:

6 0

When two hydrogen atoms bond, the positive nucleus of one atom attracts the electrons of both the atoms since the electrons are shared between the two atoms now.

You might be interested in

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers

What is the kinetic theory of matter? How does it relate to the motion of molecules?

navik [9.2K]

The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess.

3 0

3 years ago

a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to

Natalka [10]

Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

F = ma

F = (21 kg) (42.3 m/s²)

F ≈ 890 N

4 0

3 years ago

10.

myrzilka [38]

Answer:

The new period of oscillation is D) 3.0 T

Explanation:

Simple Pendulum

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

$T'=2\pi \sqrt{\frac{L'}{g}}$

$T'=2\pi \sqrt{\frac{9L}{g}}$

Taking out the square root of 9 (3):

$T'=3*2\pi \sqrt{\frac{L}{g}}$

Substituting the original T:

$T'=3*T$

The new period of oscillation is D) 3.0 T

4 0

3 years ago

A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu

ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

$\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb$

8 0

3 years ago