The vertical component of force exerted by the hi.nge on the beam will be,142.10N.
To find the answer, we need to know more about the tension.
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How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
- Let's draw the free body diagram of the system.
- To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.
- To find the answer, we have to find the tension,
- Thus, the vertical component of the force exerted by the hi.nge on the beam will be,
Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.
Learn more about the tension here:
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Nuclear fusion in the core tries to blow the star apart. Gravity holds it together. Whoever designed that system really knew what he was doing. I'm kinda grateful to him.
Answer:
Vector quantities are important in the study of motion. Some examples of vector quantities include force, velocity, acceleration, displacement, and momentum. The difference between a scalar and vector is that a vector quantity has a direction and a magnitude, while a scalar has only a magnitude. Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. A quantity which does not depend on direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude and a direction. The resulting motion of the aircraft in terms of displacement, velocity, and acceleration are also vector quantities. A vector quantity is different to a scalar quantity because a quantity that has magnitude but no particular direction is described as scalar. A quantity that has magnitude and acts in a particular direction is described as vector.
Explanation:
Answer:
The third drop is 0.26m
Explanation:
The drop 1 impacts at time T is given by:
T=sqrt(2h/g)
T= sqrt[(2×2.4)/9.8]
T= sqrt(4.8/9.8)
T= sqrt(0.4898)
T= 0.70seconds
4th drops starts at dT=0.70/3= 0.23seconds
The interval between the drops is 0.23seconds
Third drop will fall at t= 0.23
h=1/2gt^2
h= 1/2×9.81×(0.23)^2
h= 0.26m
