The radius of curvature of the proton's path while in the field is 
× 
.
b) Let R = radius curvature of protons path. Then,
relation b/w B, R, and v is: -
×
Hence, the radius of curvature of the proton's path while in the field is × .
<h3>
What do you mean by Magnetic field?</h3>
The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field. The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things.
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Answer:
This what they all been waiting for
I guess so
They been waiting for this sh,it for a long time didn't they
I'ma give it everything I got
Ayo Dougie park that X6 around the corner
Aye I'm just feeling my vibe right now
I'm feeling myself
Explanation:
So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000
<span>Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^<span>9</span>.</span>
Being that we moved the decimal six places back the answer is 5.098 x 10^6
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
I’m pretty sure it just wants you to list the property’s meaning the material and density
