The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
- Let's draw the free body diagram of the system using the given data.
- From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
- For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

- We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

- To find Ny, we need to find the tension T.
- For this, we can equate the net horizontal force.

- Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

- Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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Answer:
to have an accurate measure
Explanation:
Answer:

Explanation:
As we know that initial speed of the fall of the stone is ZERO

also the acceleration due to gravity on Mars is g
so we have

now we have

now if the same is dropped for 4t seconds of time
then again we will use above equation



Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
In physical science, there are two types of quantity: scalar and vector. While scalar quantities only include the magnitude, vector quantities include both the magnitude and the direction. Displacement is an example of vector quantities. Therefore, it includes magnitude and direction.