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AleksAgata [21]
3 years ago
8

To simulate deafness, put ear plugs or cotton balls into your ears and put your hands over your ears to block out all sound. Des

cribe this experience in terms of how it changed your perceptions of your environment and how it made you feel. What are some things you would miss if you couldn't hear? How would it change the way you live your life?
Physics
1 answer:
yulyashka [42]3 years ago
6 0

Just so you know, total deafness is quite different than partial deafness.

Partial deafness can be any number of things. It can be an inability to make out the contents of high notes (notes with high frequencies and few harmonics).

Partial deafness can be the inability to hear low notes. I only know one person who suffers from that.

Your response to speech is quite different. Usually you have to have people look right at you. Lip reading is an art. Most deaf people between what they partially hear and lip reading can make out about 1/2 of what is said. Guessing usually takes care of the rest. Course you can get everything all muddled by guessing.

It does change your way of life. The TV is often turned up high enough that people living 4 houses down can list to the 6 o'clock news when you do.

Music is hard to make out. The overtones are lost.

Earphones have very tinny sound.

People are not as enjoyable as they once were.

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George is applying a downward force of 50N to and object that has a mass of 50kg. What is the normal force (FN) of the object wh
attashe74 [19]

The normal force acting on the object is 500 N in the upward direction

<u>Explanation:</u>

As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.

So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.

But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Normal\ force = Mass \times Acceleration\ due\ to\ gravity

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then

Normal force = 50 × (-10) = -500 N

So, the normal force acting on the object is 500 N in the upward direction.

3 0
3 years ago
URGENT
olga55 [171]

If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),

(0,r) + (s cos297,s sin297) = (6,0)

now just solve for r and s.

Pls mark me as brainliest

3 0
3 years ago
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
3 years ago
If you could help me out that would be fantastic.?!!!
Volgvan
It's either C. Or B.
3 0
4 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
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