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Anastasy [175]
3 years ago
15

When a body falls to the earth, the earth also move up to meet it but the earth motion is not noticeable. why?

Physics
1 answer:
viva [34]3 years ago
7 0
It is because earths mass is very large so its movement isnt noticibale
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Chemical bonds. These bonds form as a result of the sharing or exchange of electrons between the atoms
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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

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4 years ago
What's a summary or abstract about cat eye syndrome?
Tanya [424]

Answer:

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5 0
3 years ago
According to kinetic theory, which two factors are important for a successful reaction to occur?
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A collision that meets these two criteria, and that results in a chemical reaction, is known as a successful collision or an effective collision.

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A small remote-control car with a mass of 1.65 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow
Reil [10]

Answer:

N=63.69N

Explanation:

The normal force exerted on the car by the walls of the cylinder at the bottom of the vertical circle will be such that when substracted to the weight it must give the centripetal force, since at that point on the vertical F_{cp}=N-W=N-mg

We also know that the equation for the centripetal force is:

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Which for our values means:

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