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Ksenya-84 [330]

3 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?

1 answer:

timurjin [86]3 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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An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

The cylinder height decreases by $1.2\times 10^{-4}\ m$

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3 years ago

Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously depo

amid [387]

(b) $1.54\cdot 10^{-5} C$

We can actually solve first part (b) the problem. In fact, we know that the electric field strength at the surface of the a sphere is given by

$E=\frac{kQ}{R^2}$

where

k is the Coulomb's constant

Q is the charge on the surface of the sphere

R is the radius

For this sphere, the radius is half the diameter, so

$R=\frac{43.0 cm}{2}=21.5 cm = 0.215 m$

We also know that the maximum charge is the value of charge deposited at which the electric field of the sphere becomes equal to the breakdown electric field, so

$E=3.00 \cdot 10^6 V/m$

Solving the formula for Q, we find the maximum charge:

$Q=\frac{ER^2}{k}=\frac{(3.00\cdot 10^6)(0.215)^2}{9\cdot 10^9}=1.54\cdot 10^{-5} C$

(a) $6.45\cdot 10^5 V$

The maximum potential of a charged sphere occurs at the surface of the sphere, and it is given by

$V=\frac{kQ}{R}$

where we already found at point b)

$Q=1.54\cdot 10^{-5} C$

and we know that

R = 0.215 m

Solving for V, we find:

$V=\frac{(9\cdot 10^9)(1.54\cdot 10^{-5})}{0.215}=6.45\cdot 10^5 V$

8 0

3 years ago

An emergency vehicle blowing its siren is moving

Georgia [21]

The frequency produced by the siren is 631.12 Hz

<h3>Doppler effect</h3>

The variation in frequency when a source of sound moves relative to an observer is determined by the doppler effect.

<h3>Frequency of observer</h3>

So, the frequency of the observer f' = (v ± v')f/(v ± v") where

f' = 590 Hz

f = frequency of source or siren ,

v = speed of sound = 330 m/s,

v' = speed of observer = 0 m/s (since you are stationary) and

v" = speed of source = 23 m/s

Since the source moves away from the detector, the sign in the denominator is positive and v' = 0 m/s

So, f' = (v + 0)f/(v + v")

f' = vf/(v + v")

Since, we require the frequency of the source, make f subject of the formula, we have

<h3>Frequency of siren</h3>

f = (v + v")f'/v

Substituting the values of the variables into the equation, we have

f = (v + v")f'/v

f' = (330 m/s + 23 m/s) × 590 Hz/330 m/s

f' = 353 m/s × 590 Hz/330 m/s

f' = 208270 m/sHz/330 m/s

f' = 631.12 Hz

The frequency produced by the siren is 631.12 Hz

Learn more about doppler effect here:

brainly.com/question/2169203

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They are both in a group of seven of the periodic table

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