Answer:
0.239 T
Explanation:
Applying,
F = Bvqsin∅................ Equation 1
Where F = magnetic force, B = magnetic Field, q = charge of a proton, v = velocity of proton, ∅ = angle between the velocity and the magnetic field.
make B the subject of the equation
B = F/(vqsin∅)................. Equation 2
From the question,
Given: F = 1.15×10⁻¹³ N, v = 3.0×10⁶ m/s, ∅ = 90°(perpendicular)
Constant: q = 1.602 x 10⁻¹⁹ C
Substitute into equation 2
B = 1.15×10⁻¹³ /(3.0×10⁶×1.602 x 10⁻¹⁹×sin90°)
B = 1.15×10⁻¹³/(4.806×10⁻¹³)
B = 0.239 T.
Hence the magnetic field = 0.239 T
Answer: The acceleration that is directed radially towards the center of the circle having a magnitude equal to the square of the speed of the body along the curve is divided by the total distance from the center of the circle to the moving body.
Explanation:
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Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change
Δe=0.578 kJ/kg
I believe it’s tackling but I’m not quite sure
