What is the last planet discovered?
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As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that
now from the above equation
so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by
here we know that angle is
now we have
so net force on q3 is 0.46 N vertically upwards along +Y axis
Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)