Which would be the best example to demonstrate the principle of conservation of energy? A) a stone is crushed B) light bends aro
2 answers:
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Answer:
Explanation:
Given data:
Mass of the man,
Total mechanical energy,
Height,
Suppose there is no external force acting on the man. In this situation, the total mechanical energy (kinetic + potential) will remain steady.
Let the speed of the man at 2.6 m be <em>v</em>.
Thus,
Answer:
Explanation:
Given that,
The ray makes an angle of 32 degrees with respect to the normal of the surface.
The refractive index of air,
The refractive index of water,
Snell's law is given by :
So, option (4) is correct. Hence, this is the required solution.
Answer:
given,
Probability of student studying math P(M)=
Probability of student studying physicsP(P) =
Probability of student studying both math and physics together P(M∩P) =
a) student took mathematics or physics
P(M∪P) = P(M) + P(P) - P(M∩P)
= 0.45 + 0.61 - 0.25
= 0.81
b) student did not take either of the subject
P((M∪P)') = 1 - 0.81
= 0.19
c) Student take physics but not mathematics
P(P∩M') = P(P) - P(P∩M)
= 0.61 - 0.25
= 0.36
studying physics and mathematics is not mutually exclusive because we can study both the subjects.
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
