Find the equivalent resistance of this parallel circuit with two strands.
1 answer:
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Refer to the diagram shown below.
Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.
The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)
Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N
The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N
If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²
Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.
Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.
The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)
Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N
The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N
If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²
Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.
Answer:
Explanation:
The tension in the cable would be then equal to the gravitational force that keeps the Moon in circular orbit around the Earth. So, we just need to calculate the magnitude of this force, which is given by:
where:
is the gravitational constant
is the mass of the Moon
is the mass of the Earth
is the distance between the Earth and the Moon
Substituting these numbers into the formula, we find: