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A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10ms–1. There is a uniform horizontal electric field of 104N/C.
- a)the time of flight of the particle is 2sec
- b)the horizontal range of the particle is 10m
- c)the horizontal range of the particle is 0m
- d)the maximum height reached is 5m
Correct answer is option 'A,B,D'. Can you explain this answer?
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A particle of mass 2kgand charge 1mCis projected vertically with a vel...
The correct answers are: the horizontal range of the particle is 10m, the time of flight of the particle is 2sec, the maximum height reached is 5m
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A particle of mass 2kgand charge 1mCis projected vertically with a vel...
Given data:
Mass (m) = 2kg
Charge (q) = 1mC = 1 x 10^-3 C
Initial velocity (u) = 10m/s
Electric field (E) = 104 N/C
Acceleration due to gravity (g) = 9.8 m/s^2
a) Time of flight:
Using the equation of motion:
v = u + at, where v = final velocity, u = initial velocity, a = acceleration, t = time
The vertical acceleration of the particle is due to gravity and electric field, so a = g + (qE)/m
Substitute the values to get a = 9.8 + (1 x 10^-3 x 104)/2 = 9.8 + 0.052 = 9.852 m/s^2
Now, using v = u + at, we get 0 = 10 - 9.852t
Solving for t, we get t = 10/9.852 = 1.015 seconds
b) Horizontal range:
Using the equation:
Range (R) = u*t
Substitute the values to get R = 10 * 1.015 = 10.15 meters
c) Horizontal range:
As the particle is projected vertically, there will be a horizontal range covered by the particle. Therefore, the horizontal range cannot be zero.
d) Maximum height reached:
Using the equation:
Maximum height (H) = (u^2)/(2g)
Substitute the values to get H = (10^2)/(2*9.8) = 5.102 meters
Therefore, the correct answers are A, B, and D.
Mass (m) = 2kg
Charge (q) = 1mC = 1 x 10^-3 C
Initial velocity (u) = 10m/s
Electric field (E) = 104 N/C
Acceleration due to gravity (g) = 9.8 m/s^2
a) Time of flight:
Using the equation of motion:
v = u + at, where v = final velocity, u = initial velocity, a = acceleration, t = time
The vertical acceleration of the particle is due to gravity and electric field, so a = g + (qE)/m
Substitute the values to get a = 9.8 + (1 x 10^-3 x 104)/2 = 9.8 + 0.052 = 9.852 m/s^2
Now, using v = u + at, we get 0 = 10 - 9.852t
Solving for t, we get t = 10/9.852 = 1.015 seconds
b) Horizontal range:
Using the equation:
Range (R) = u*t
Substitute the values to get R = 10 * 1.015 = 10.15 meters
c) Horizontal range:
As the particle is projected vertically, there will be a horizontal range covered by the particle. Therefore, the horizontal range cannot be zero.
d) Maximum height reached:
Using the equation:
Maximum height (H) = (u^2)/(2g)
Substitute the values to get H = (10^2)/(2*9.8) = 5.102 meters
Therefore, the correct answers are A, B, and D.
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A particle of mass 2kgand charge 1mCis projected vertically with a vel...
The correct answers are: the horizontal range of the particle is 10m, the time of flight of the particle is 2sec, the maximum height reached is 5m
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A particle of mass 2kgand charge 1mCis projected vertically with a velocity 10ms–1. There is a uniform horizontal electric field of 104N/C.a)the time of flight of the particle is 2secb)the horizontal range of the particle is 10mc)the horizontal range of the particle is 0md)the maximum height reached is 5mCorrect answer is option 'A,B,D'. Can you explain this answer?
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A particle of mass 2kgand charge 1mCis projected vertically with a velocity 10ms–1. There is a uniform horizontal electric field of 104N/C.a)the time of flight of the particle is 2secb)the horizontal range of the particle is 10mc)the horizontal range of the particle is 0md)the maximum height reached is 5mCorrect answer is option 'A,B,D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle of mass 2kgand charge 1mCis projected vertically with a velocity 10ms–1. There is a uniform horizontal electric field of 104N/C.a)the time of flight of the particle is 2secb)the horizontal range of the particle is 10mc)the horizontal range of the particle is 0md)the maximum height reached is 5mCorrect answer is option 'A,B,D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 2kgand charge 1mCis projected vertically with a velocity 10ms–1. There is a uniform horizontal electric field of 104N/C.a)the time of flight of the particle is 2secb)the horizontal range of the particle is 10mc)the horizontal range of the particle is 0md)the maximum height reached is 5mCorrect answer is option 'A,B,D'. Can you explain this answer?.
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