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3 years ago

Please help only eight questions C:

Physics

2 answers:

ikadub [295]3 years ago

7 0

1. 2 is the answer

2. 3 is the answer

3. 1 is the answer

4. 3 is the answer

hope this helps

Strike441 [17]3 years ago

5 0

1 2 answer

2 3 answer

3 1 answer

4 3 answer

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What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

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Answer:

charge will be equal to $2.03\times 10^{-5}C$

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity $g=9.8m/sec^2$

Electric field E = 700 N/C

Electric force will be equal to $F=qE$ , here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

$q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C$

So charge will be equal to $2.03\times 10^{-5}C$

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