The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,




Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
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Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
Answer:
i'm pretty sure its B but i may be wrong if you dont wanna take the chance wait for someone
Explanation:
The correct answer Is B-balanced
4000 seconds
Explanation:
speed = distance / time
0.0004m/s = 1.6m / time
Subject time
time = 1.6 / 0.0004
time = 4000 seconds.
Hope this helps. Mark as brainliest if possible. tks