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4 years ago

When deciding antenna types, why might the use of an omnidirectional antenna be inadvisable?

Physics

1 answer:

d1i1m1o1n [39]4 years ago

7 0

Answer:

Omnidirectional antennas are inadvisable to be used because when oriented vertically they radiate radio power equally in all horizontal directions, but heir radio power varies with the angle to the axis, reducing to zero at the axis.

Explanation:

Omnidirectional antennas radiates equal radio power in all directions perpendicular to an axis, with power varying with the elevation angle. This means that their power declines to zero on the axis. This is unlike the isotropic antennas that radiates equal power in all directions, without any reduction in radio power with angle to the axis. Omnidirectional antennas are mostly used for radio broadcasting antennas in mobile devices that use radio such as cell phones, FM radios, walkie-talkies, wireless computer networks etc.

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In the picture above, what is the period of the pendulum?<br> (Remember to include a unit)

-BARSIC- [3]

69 i agree with her hope this helps

6 0

3 years ago

The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time b

alekssr [168]

Answer:

a - As long as the time between 2 events is reconcilable with a light signal, the time between the events, in that frame, can be determined.

8 0

3 years ago

Read 2 more answers

An automobile rounds a curve of radius 50.0 m on a flat road.

bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve = 50m

Centripetal acceleration = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

It is given as;

a = $\frac{v^{2} }{r}$

a is the centripetal acceleration

v is the speed

r is the radius

Now insert the parameters and find v;

v² = ar

v² = 3.92 x 50 = 196

v = √196 = 14m/s

6 0

3 years ago

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy.

statuscvo [17]

Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.

According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J

Minus sign indicates that work is done by the system.

7 0

3 years ago

Read 2 more answers

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago