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antiseptic1488 [7]

2 years ago

10

Determine the field strength, E, experienced by a test charge, q, if a charge of 7.0 × 10-5 coulombs is placed on q and a force

of 5.2 N is applied.

2 answers:

kkurt [141]2 years ago

7 0

Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=

Evgesh-ka [11]2 years ago

5 0

E = F/q

= 5.2 N / (7* 10⁻⁵) C

= 7.429 * 10⁴ N/C

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xz_007 [3.2K]

Answer:

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Explanation:

We can use the following kinematics equation

$v_f^2-v_i^2=2 \ a \ d$

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The force will be tripled, the force is:

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in 1D

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Now, for the original problem, we have

$(25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'$

$(25 \frac{m}{s}))^2=2 \ a' \ d'$

$625 \frac{m^2}{s^2}=2 \ a' \ d'$

For the second problem, we have

$(v_f})^2-(v_i)^2=2 \ a'' \ d''$

starting from the rest, we have the same initial velocity.

$(v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''$

$(v_f})^2=2 \ a'' \ d''$

As the force is tripled, we have:

$F'' = 3 F'$

$m'' \ a'' = 3 m' \ a'$

But the mass its the same, so

$m' \ a'' = 3 m' \ a'$

$a'' = 3 \ a'$

So the acceleration its also tripled.

$(v_f})^2=2 \ (3 * a') \ d''$

$(v_f})^2=3 ( 2 \ ( a' \ d'' )$

As the distance traveled by the arrow must also be the same, we have:

$(v_f})^2=3 ( 2 \ ( a' \ d' )$

$(v_f})^2=3 (625 \frac{m^2}{s^2})$

$v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}$

$v_f= \ sqrt{3} 25 \frac{m}{s}$

$v_f= 43.30 \frac{m}{s}$

And this will be the speed from the arrow leaving the bow.

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Learn more about velocity:

brainly.com/question/5248528

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