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Vadim26 [7]
3 years ago
7

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

f 0.3m apart in a vacuum.​
Physics
1 answer:
adell [148]3 years ago
7 0

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

        F =k \frac{q_1q_2}{r^2}

In this case they indicate that the load is of equal magnitude

       q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

       q = \sqrt{\frac{F \ r^2}{k} }

we calculate

        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

        q = \sqrt{7 \ 10^{-12} }Ra 7 10-12

        q = 2.65 10⁻⁶ C

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3 years ago
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
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Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

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T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

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3 years ago
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E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

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3 years ago
Topic Gravitational force amd firld strength.. help me please
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The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

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The solution with the negative square root is negative, so we throw it out. The other is the one we want,

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