3 years ago

VAMDNVABNATNAMDTEMCYSPLDDNHNMQTYYque es bandaaa

Physics

2 answers:

Sliva [168]3 years ago

8 0

Im sorry but i dont understand

kvasek [131]3 years ago

3 0

De que hablas bandaaaaaa

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2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What

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3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

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Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago

VAMDNVABNATNAMDTEMCYSPLDDNHNMQTYYque es bandaaa​

VAMDNVABNATNAMDTEMCYSPLDDNHNMQTYYque es bandaaa