Draw the force vector starting at the black dot. The location, orientation, and length of the vector will be graded. You can mov
e the vectors F⃗ 1F→1 and F⃗ 2F→2 to construct the required vector, but be sure to return them into their initial positions before submitting the answer.
1 answer:
Answer:
F3 is the equilibrant force equal in magnitude to the resultant between F1 and F2 but opposite in direction to it.
Explanation:
Given the diagram, the force F3 to make the body remain at rest or in equilibrium is the equilibrant force, this force is equal in magnitude to the resultant Fr between F1 and F2 but opposite in direction to it.
See attachment for diagram of forces.
The resultant force;
Fr =√ (F1)² + (F2)²...(1) [diagram a]
Therefore the length of F3 is Fr
F3 = -Fr
The direction (diagram b) of the resultant force Fr is given by
∆ = acrtan[(F1/F2)]
The direction of F3 is [90 + arctan(F1/F2)]
As seen in the diagram (d), the location of the force is in the fourth quadrant.
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Answer:
e = 1.21 mV
Explanation:
given,
length of rod = 10 m
height of drop = 4.89 m
Earth’s magnetic field = 12.4 µT
acceleration of gravity = 9.8 m/s²
velocity of the beam
v = 9.79 m/s
emf of the beam
e = B l v
e = 12.4 x 10⁻⁶ x 9.79 x 10
e = 1.21 x 10⁻³ V
e = 1.21 mV
Answer:
2.67kg
Explanation:
The maximum velocity, of a body experiencing simple harmonic motion is given by equation (1);
where is the angular velocity and A is the amplitude.
The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);
where k is the force constant of the spring and m is the loaded mass.
We can make the subject of formula in equation (1) as follows;
We then combine equations (2) and (3) as follows;
According to the problem, the following are given;
We then substitute these values into equation (4) and solve for the unknown mass m as follows;
Squaring both sides, we obtain the following;