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Novosadov [1.4K]
3 years ago
13

Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

the beam of electrons(beta particles)?Medal will be given to best answer
Physics
1 answer:
wlad13 [49]3 years ago
4 0
It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.
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1. Why did Asteroid 2019 OK cause concern among scientists? * The asteroid was not detected until it was extremely close to Eart
AveGali [126]

Answer:

<u>The asteroid was not detected until it was extremely close to Earth. </u>

Explanation:

According to data from NASA, the Asteroid named 'Astriod 2019 OK', was detected when it was extremely close to earth with just about an estimated distance of  73,000 kilometers  (45,000 miles) from the Earth.

Scientists were concerned at the proximity of this space object to the Earth before it was discovered, and it brought about a cause of concern that since it was not extremely large (estimated 57 to 130 meters wide) it creates a potential for other smaller asteroids to escape detection and struck the earth.

7 0
2 years ago
How far does a wave travel in three cycles?
eduard
It will be 3 wavelengths because 1 cycle = 1 wavelength. 
6 0
3 years ago
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/
LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\

we isolate th (needed to reach the maximum height hmax)

th = \frac{v_{0}*sin(\alpha)}{g}

The formula describing vertical distance is:

y = Vy * t-g* t^{2} / 2

So, given y = hmax and t = th, we can join those two equations together:

hmax = Vy* th-g*th^{2}/2

hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)

if we launch a projectile from some initial height h all you need to do is add this initial elevation

hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)

hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft

6 0
3 years ago
30 points!
vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc)  items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0
2 years ago
The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is
astraxan [27]

Complete question:

Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.

Answer:

The magnitude of this field is 826 N/C

Explanation:

Given;

The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m

PEsinθ = T

where;

E is the magnitude of the electric field

P is the dipole moment

First, we determine the magnitude dipole moment;

Magnitude of dipole moment = q*r

P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m

Finally, we determine the magnitude of this field;

E = \frac{T}{P*sin(\theta)}=  \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C

E = 826 N/C (in three significant figures)

Therefore, the magnitude of this field is 826 N/C

6 0
3 years ago
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