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Wewaii [24]
3 years ago
15

A 0.6kg ball accelerated at 55 m/s​ 2​ . What force was applied?

Physics
1 answer:
adell [148]3 years ago
7 0

Answer: 33N

Explanation:

Newton's second Law of motion F = ma

F = (0.6)(55) = 33N

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What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes
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The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

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  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

  • E is electric field
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F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
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Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

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