Answer:
Explanation:
Let mass and velocity of proton be m and v .
1/2 m v² = 59 x 10⁶ e V
= 59 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 94.4 x 10⁻¹³ J
mv² = 188.8 x 10⁻¹³ J
v² = 188.8 x 10⁻¹³ / m
= 188.8 x 10⁻¹³ / 1.67262 x 10⁻²⁷
= 112.8768 x 10¹⁴
v = 10.62 x 10⁷ m / s
In circular path of proton , magnetic force equals centripetal force .
m v² / r = B q v , B is magnetic field , q is charge on proton , r is radius of circular path .
188.8 x 10⁻¹³ / 5.8 x 10¹⁰ = B x 1.6 x 10⁻¹⁹ x 10.62 x 10⁷
B = 1.9157 x 10⁻¹¹ T.
0 Newtons, since it's at a constant speed. Hope this helps, and Brainliest answer would be appreciated!
Answer:24
Explanation:
First you need to calculate the acceleration, for this use can use the following formula Force (F) = Mass (M) x Acceleration (A). If we extract A out of this, you can get the following equation A = F/M = 4/0.5 = 8 m/s^2. So, the object accelerates each second by 8 m/s, if the object does this for 3 second, the velocity would be 8 x 3 = 24
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
It would make them different. they wouldn't be the same.