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Thepotemich [5.8K]
3 years ago
15

Heyy i don't understand this question may u help me pls?​

Physics
2 answers:
Elan Coil [88]3 years ago
8 0

Answer:

What question?

Explanation:

Oksana_A [137]3 years ago
8 0
What’s the question
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The observation location A is located at < 0.112, 0.247, 0 > m. The conventional current running through the wire is 7.5 a
xenn [34]

Answer:

Positive direction

Explanation:

We know that the direction of flow of current is always opposite the direction of the flow of electrons. Assuming that the electrons move in negative direction then the current will flow in the positive direction.

5 0
3 years ago
What is a Super Massive Black Hole?
Serggg [28]

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

4 0
3 years ago
A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1
liubo4ka [24]

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

4 0
3 years ago
Important nuclei of the indirect (multineuronal) system that receive impulses from the equilibrium apparatus of the inner ear an
Lera25 [3.4K]

Answer:

B. Vestibular nuclei

Explanation:

The nerve information generated by the vestibular receptors travels through the vestibular portion of the eighth pair that penetrates the brain stem at the level of the brain stem bridge. At this level there are four vestibular nuclei, which receive the synapses of these axons, coming from the ridges and macules. The semicircular ducts predominantly terminate in the superior and medial nuclei. While the fibers coming from the macules end on the lateral, medial and inferior nuclei. Some fibers of the eighth pair end in the flocculonodular lobe of the cerebellum, <u>these connections play an important role in controlling posture and balance.</u>

From the vestibular nuclei, two bundles of fibers descending to the spinal cord originate from the medial and lateral vestibular spinal bundles and a bundle of fiber that rises in the brain stem that participates in the coordination of eye movements, the medial longitudinal fascicle, which participates in Rotational nystagmus This system also participates in an important way in the control of some ocular movements by the fibers that it contributes to the medial longitudinal fascicle, which is a structure that interconnects the motor nuclei of the extrinsic muscles of the eyeballs VI or abdicens nucleus (abductor) on one side and IV or pathetic nucleus (trochlear) and III or nucleus of the common ocular motor (oculomotor) on the opposite side.

6 0
3 years ago
A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football
KengaRu [80]

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

p_{i} = p_{f}  

m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}  (1)

Where:

m_{a}: is the mass of the first football player = 110 kg

m_{b}: is the mass of the second football player = 90 kg

v_{a_{i}}: is the initial velocity of the first football player = 5.0 m/s

v_{b_{i}}: is the initial velocity of the second football player = 0 (he is at rest)

v_{a_{f}}: is the final velocity of the first football player = 0 (he stops after the impact)

v_{b_{f}}: is the final velocity of the second football player =?

By solving equation (1) for v_{b_{f}} we have:

110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}

v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!

8 0
3 years ago
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