Im so confused can someone help. I’ll give you BRAINLIETS
1 answer:
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1. Elastic potential energy (D. EEl)
In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:
where k is the spring constant and x is the compression of the spring.
2. Gravitational potential energy (C. Eg)
In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:
where m is the mass of the toy and g is the gravitational acceleration.
3. Gravitational potential and kinetic energy (A. Eg and EK)
In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by
Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.
4. Gravitational potential energy (C. Eg)
The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:
where m is the mass of the jumper.
5. This exercise has the same text of the previous one.
Answer:
tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N
Explanation:
The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.
The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem
For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.
We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis
∑ Fx = 0
Fx -T =0
∑ Fy =0
Fy -W =O W = m g
∑τ =0
-T dy + W dx =0
For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system
dy = L sin 17
dx = (L/2) cos 17
Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation
-T L sin 17 + mg (L / 2) cos 17 = 0
- T sin 17 + mg/2 Cos 17 =0
We write the three equations together
Fx- T = 0
Fy -W = 0
-T sin 17 + (m g / 2) cos 17 = 0
a) With the third equation we can find the wire tension
T = m g / 2 cos 17 / sin17
T = 6.5 9.8/2 cotan 17
T = 104 N
b) We use the first equation to find Fx
Fx- T =0
Fx = T = 104 N
c) We use the second equation to find Fy
Fy = W = m g
Fy = 6.5 9.8
Fy = 63.7 N
