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svet-max [94.6K]
3 years ago
7

1. Line segment AC touches the circle at a single point B. Line segment OB extends through the center of the circle.

Physics
1 answer:
adelina 88 [10]3 years ago
8 0
AC is a tangent to the circle

A tangent is perpendicular to the diameter of the circle, so B is 90 degrees

Radius is half the diameter, so 3.5*2 = 7m

Area = (pi)(r^2), so (pi)*3.5^2 = 38.48m^2


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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

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.

                           N_x=86.62N

  • 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.

                           N_y=F_V=mg-Tsin59\\

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

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

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

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

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

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

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2 years ago
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andrezito [222]

Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

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A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte
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Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

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Also,

mass of sack = 500

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velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

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light waves. the other waves like gamma rays or infrared waves or radio waves are all not visible to the eye. light is the only thing out of those you can see.

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