Gravity equation

1. Three particles each with a mass of 1 kg are at the vertices of an equilateral triangle whose sides are 1 m long. How large is the gravitational force experienced by each point particle (in G)?

SolutionGravity equation 1

The magnitude of the gravitational force experienced by one of the particles.

F12 = G (m1)(m2) / r2 = G (1)(1) / 12 = G/1 = G

F13 = G (m1)(m3) / r2 = G (1)(1) / 12 = G/1 = G

Resultant gravitational force at point 1:

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Electric field equation

1. A conducting ball with a radius of 10 cm has an electric charge of 500 μC. Points A, B, and C lie in line with the center of the ball at a distance of 12 cm, 10 cm and 8 cm respectively from the center of the ball. Calculate the electric field strength at points A, B, and C!

Known:Electric field equation 1

The radius of the conducting ball (R) = 10 cm = 0.1 m

Electric charge (q) = 500 μC = 500 x 10-6 C

rA = 12 cm = 0,12 m

rB = 10 cm = 0,1 m

rC = 8 cm = 0,08 m

Coulomb constant (k) = 9 x 109

Wanted: The electric field strength at point A (EA), at point B (EB) and at point C (EC)


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Spring constant equation

1. A spring in free suspension has a length of 10 cm. At the free end, a 200 gram weight is suspended so that the length of the spring is 11 cm. If g = 10 m/s2, what is the spring force constant?


The initial length of the spring (y1) = 10 cm = 0.10 m

The final length of the spring (y2) = 11 cm = 0.11 m

Spring length change (Δy) = 0.11 – 0.10 = 0.01 meter

The mass of the load (m) = 200 grams = 0.2 kg

Load weight (w) = m g = (0,2)(10) = 2 Newtons

Wanted: Spring constant (k)


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Potential difference equation

1. An electric charge is moved in a homogeneous electric field with a force of 2√3 N a distance of 20 cm. If the direction of the force is at an angle of 30o to the displacement of the electric charge, what is the difference in the electric potential energy at the initial and final positions of the electric charge.


Force (F) = 2√3 N

Distance (s) = 20 cm = 0.2 m

Angle (θ) = 30o

Wanted: Electric potential difference


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Impulse equation

1. A car with a mass of 250 kg is moving with a speed of 72 km/hour, then accelerated with a constant force so that in 5 seconds its speed becomes 80 km/hour. Determine the impulse for 5 seconds


The mass of the car (m) = 250 kg

Initial speed (vo) = 72 km/h = 20 m/s

Final speed (vt) = 80 km/h = 22 m/s

Time interval (t) = 5 seconds

Wanted: Determine impulse (I)


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Friction force equation

1. Block A 3 kg is placed on the table and then tied to a rope that is connected to stone B = 2 kg through a pulley as shown. The mass and friction of the pulleys are neglected. Acceleration due to gravity g = 10 m/s2. Determine the acceleration of the system and the tension in the rope if:

a) smooth tableFriction force equation 1

b) rough table with a coefficient of kinetic friction of 0.4


The mass of block A (mA) = 3 kg

The mass of rock B (mB) = 2 kg

Acceleration due to gravity (g) = 10 m/s2

Weight of block A (wA) = m g = (3)(10) = 30 Newton

Weight of rock B (wB) = m g = (2)(10) = 20 Newton

Wanted: The acceleration of the system (a) and the tension in the rope (T)


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Normal force equation

1. A block has a mass of 5 kg. If g = 10 m/s2, determine:

a) the weight of the block

b) the normal force if the block is placed on a flat plane

c) the normal force if the block rests on an inclined plane that forms an angle of 30o to the horizontal


The mass of the block (m) = 5 kg

Acceleration due to gravity (g) = 10 m/s2

Wanted: w, N on the plane and N on the incline


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Rope tension equation

1. The picture below shows three blocks, namely A, B and C which are located on a smooth horizontal plane. If mass A = 1 kg, mass B = 2 kg and mass C = 2 kg and F = 10 N, then determine the ratio of the tension in the rope between A and B to the tension in the rope between B and C.

Known:Rope tension equation 1

The mass of A (mA) = 1 kg

Mass B (mB) = 2 kg

The mass of C (mC) = 2 kg

Tensile force (F) = 10 N

Wanted: TAB : TBC


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