Answer:
The magnitude and direction of the force between the two charges is 497.66 N in opposite direction.
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 1.4×10¯³ C
Charge 2 (q₂) = 9.1×20¯⁴ C
Distance apart (r) = 4.8 m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The magnitude of the force between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 1.4×10¯³ × 9.1×20¯⁴ / 4.8²
F = 11466 / 23.04
F = 497.66 N
Thus, the magnitude of the force is 497.66 N.
Since both charges has the same sign (i.e they both positive), the force between them will be a force of repulsion. Hence the force will be in opposite direction.
Scales for weight
Any beaker to measure the volume of liquid displaced
I think it’s the first one
Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
