from the calculations, we
cansay that (V.F) almost equal to Zero.
The angle is nearly 90 degree. It means V and F are perpendicular to each other.
Thank you Exercise 27.8 A particle with charge 6.00 nC is moving in a uniform magnetic...
A particle with charge − 5.60 nC is moving in a uniform magnetic field B⃗ =−( 1.26 T )k^. The magnetic force on the particle is measured to be F⃗ =−( 4.00×10−7 N )i^+( 7.60×10−7 N )j^. Calculate the scalar product v⃗ ⋅F⃗ .
A particle with a charge of q = -5.60 nC is moving in a uniform magnetic field of B⃗ =( -1.23 T ) k^. The magnetic force on the particle is measured to be F⃗ =( −7.60×10−7 N )j^. Part B Calculate vx, the x component of the velocity of the particle. Express your answer in meters per second. m/s
Part B A particle with charge – 5.80 nC is moving in a uniform magnetic field B= -( 1.30 T )k. The magnetic force on the particle is measured to be F=-( 4.00x10-7 N )i+( 7.60x10-7N)j. Calculate the x-component of the velocity of the particle. EVO AO R O P ? Submit Request Answer Part C Calculate the y-component of the velocity of the particle. VALD * o ? Part D A particle with charge – 5.80 nC is moving...
A particle with charge − 5.40 nC is moving in a uniform magnetic field B⃗ =−( 1.28 T )k^. The magnetic force on the particle is measured to be F⃗ =−( 4.00×10−7 N )i^+( 7.60×10−7 N )j^. Part A Part complete Are there components of the velocity that are not determined by the measurement of the force? Are there components of the velocity that are not determined by the measurement of the force? yes no SubmitPrevious Answers Correct Part B...
Exercise 27.8 A particle with charge -5.90 ncis moving in a uniform magnetic seld -(1.30 T)k The magnetic force on the partcle is measured to be Part B Calculate the x component of the velocity of the particle. Submit My Answers Give Up Part C Calculate the y component of the velocity of the particle. My Answers Give Up Part D Calculate the scalar product G.F m/s m/s m/s. N
A particle with charge – 5.20 nC is moving in a uniform magnetic field B= -(1.24 T )k. The magnetic force on the particle is measured to be F = -(3.00x10-7N)+(7.60x10-7N) Part B Calculate the x-component of the velocity of the particle. I ALP * O a ? V2 = Submit Request Answer Part C Calculate the y-component of the velocity of the particle. IVO AQ o ? ปี Submit Request Answer A particle with charge – 5.20 nC is...
A particle with a charge of q = -5.20 nC is moving in a uniform magnetic field of B with sole component B, = -1.28 T. The magnetic force on the particle is measured to be F with sole component Fy = -7.60x10-7 N Calculate Vz, the x component of the velocity of the particle. Express your answer in meters per second. View Available Hint(s) ΡπΙ ΑΣφ ? v_v_x = 144.2 m/s Submit Previous Answers X Incorrect; Try Again; 3...
A particle with charge - 5.00 nC is moving in a uniform magnetic field B=-(1.24 T ). The magnetic force on the particle is measured to be =-(3.80-10-7N)2+(7.60-10-7N)). Part B Calculate the 1-component of the velocity of the particle. Express your answer in meters per second. | ΑΣφ ? m/s Submit Request Answer Part C Calculate the y-component of the velocity of the particle. Express your answer in meters per second. | ΑΣφ ? m/s Submit Request Answer Part D...
Constants A particle with charge 5.60 nC is moving in a uniform magnetic field B(1.27 T)k, The magnetic force on the particle is measured to be F = -(4.00x10-7 N )? + ( 7.60x10o-7 N. Part A Are there components of the velocity that are not determined by the measurement of the force? yes no Previous Answers Submit Correct Part B Calculate the x-component of the velocity of the particle. Πνα ΑΣφ ? Part B Calculate the x-component of the...
5.A particle of charge +2.00 x 103 C is moving with velocity v (1000 m/s)j through a uniform magnetic field. The magnetic force on the particle is F = (4.00 N)i-(6.00 N)k. What is B, the z component of the magnetic field (in T)? (A) 1.00 (B) 1.50 (C) 4.00 (D) 2.00 (E) 3.00 (F) 5.00