n a region where there is a uniform electric field that is upward and has magnit
ID: 1643734 • Letter: N
Question
n a region where there is a uniform electric field that is upward and has magnitude 3.80×104 N/C , a small object is projected upward with an initial speed of 2.12 m/s . The object travels upward a distance of 6.98 cm in 0.200 s.
Part A
What is the object's charge-to-mass ratio q/m (magnitude and sign)? Assume g = 9.80 m/s2, and ignore air resistance.
Hello, thanks for your help in advance.
This is for practice for my exam. He requires that we set up an energy diagram before any work done. He also asks us to work all the equations with variables. Once they're where we need them then he asks that we simply plug and chug.
Could you help me with this, please?
n a region where there is a uniform electric field that is upward and has magnitude 3.80×104 N/C , a small object is projected upward with an initial speed of 2.12 m/s . The object travels upward a distance of 6.98 cm in 0.200 s.
Part A
What is the object's charge-to-mass ratio q/m (magnitude and sign)? Assume g = 9.80 m/s2, and ignore air resistance.
Hello, thanks for your help in advance.
This is for practice for my exam. He requires that we set up an energy diagram before any work done. He also asks us to work all the equations with variables. Once they're where we need them then he asks that we simply plug and chug.
Could you help me with this, please?
Explanation / Answer
From the given question,
Electric field(E)=3.80×104 N/C
Initial speed(u)=2.12 m/s
Distance travelled(x)=6.98cm=0.0698m
Time(t)=0.2 sec
From equations of kinematics,
s=ut+(1/2)at2
0.0698=(2.12)(0.2)+ (1/2)(a)(0.2)2
0.0698=0.424+0.02a
a=-17.71
F=ma
qE=ma
q/m=a/E
= -17.71/3.8 x 104
=- 4.66 x 10-4
q/m= - 4.66 x 10-4
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