Give two examples of contact force and non contact force

Physics

1 answer:

Crank2 years ago

5 0

Answer:

Explanation:

The two examples of contact forces are frictional force and applied force. The two examples of noncontact forces are gravitational force and magnetic force.

You might be interested in

The critical angle for a transparent

mamaluj [8]

Answer:

1.58901572907

Explanation:

n = 1 / sin(c)

$n=\frac{1}{sin(c)}$

4 0

2 years ago

Why is important the electromagnetism, diff usages, and how will be our World without it

lawyer [7]

Answer:

Electromagnetism has important scientific and technological applications. It is used in many electrical appliances to generate desired magnetic fields. It is even used in a electric generator to produce magnetic fields for electromagnetic induction to occur. It has many more technological applications including MRI scanning (magnetic resonance imaging) and electric bells.

Explanation:

Electromagnets are widely used as components of other electrical devices, such as motors, generators, electromechanical solenoids, relays, loudspeakers, hard disks, MRI machines, scientific instruments, and magnetic separation equipment.

7 0

3 years ago

Which does not contain a lens?

statuscvo [17]

C a glass eye is my best guess because its all glass and a fake eye not actually working so thats my theroy.

8 0

3 years ago

In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular moti

AleksandrR [38]

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

$F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\$

$But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz$

6 0

3 years ago

What is meant by saying that a wave has a low frequency?

OverLord2011 [107]

The answer is B. The wave is very long.

3 0

4 years ago