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Ksenya-84 [330]
2 years ago
7

A guy wire helping to stabilize a transmitting tower 500 m high makes in angle of 50° with the ground. In a strong wind, the tow

er pulls The wire with a force of 1500.N. The anchorage of the wire is Secure against any lateral pull, but it cannot withstand in an upward pull of more than 1200.N. Does the wire come loose?
Physics
1 answer:
timurjin [86]2 years ago
6 0

They give us the cable tension t = 1500 newton. We assume that the mass of the cable is negligible compared to that of the tower.

We have a force t of 1500 Newton. This force has a vertical component on the y axis and a horizontal component on the x axis.

Of these two components of force, we are especially interested in calculating is the magnitude of the vertical component.

If the angle that it forms with the ground is 50 °, then the vertical component of the force is:

Fy = 1500sin (50)

Fy = 1149 N. <1200 N

The wire will not be loose

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The two spheres pictured above have equal densities and are subject only to their mutual gravitational attraction. Which of the
kiruha [24]

Answer:

Gravitational force

Explanation:

If two spheres have equal densities and they are subject only to their mutual gravitational attraction. We need to say that the quantities that must have the same magnitude for both spheres. So, the correct option is (E) i.e. gravitational force.

It is because of Newton's third law of motion. It states that the force due to object 1 to object 2 is same as force due to object 2 to object 1. The two forces act in opposite direction.  

Hence, the correct option is (E) "Gravitational force".                        

8 0
2 years ago
1. What is the intensity of a sound when an observer is 10.0 m from the speaker
Nana76 [90]

Answer:

About 133 db.

Explanation:

Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of  the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.

I = 200 w / 10 m^2 = 20 w per square meter  

I sub 0 = 10^-12 w per square meter

SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db

Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D

8 0
3 years ago
Which statements best describe displacement? Check all that apply.
xxMikexx [17]

Answer:

the last one, the third one, and the first one.

8 0
2 years ago
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The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
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