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2 years ago

Which type of communications equipment functions as a radio receiver and searches across several frequencies?

1 answer:

7 0

A <u>scanner</u> is a type of communications equipment that functions as a radio receiver and searches across several frequencies.

A scanner is a kind of a radio receiver that has the ability to receive multiple signals.

There are three modes which a scanner uses for acting as a radio receiver. The scan mode of the radio receiver constantly changes frequencies that helps in transmissions. There is also a manual scan mode that allows the users to search for their interested frequencies. The search mode allows the users to search through two sets of frequencies.

A scanner is a type of communication equipment that is easy to use with various features such as the volume, numeric keypad, trunk tracking etc.

To learn more about scanners, click here:

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When a solid uniform sphere is spinning about an axis of rotation through its center, its rotational kinetic energy is K and mom

lawyer [7]

Answer:

uh.

Explanation:

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2 years ago

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The forces on a car are balanced.state and explain the resultant force on the car

Vanyuwa [196]

there will no resultant force

Explanation:

this is because if the forces are balanced on opposite direction. then they cancel each other out

5 newton's ---------> <--------- 5 newton's

then both forces will cancel each other out as a result there is no resultant force and the newton's laws states that if there is no resultant the object will continue in its state of rest (remains there) or it will in continue in its uniform motion in a straight line.

I hope you understand,

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3 years ago

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0

3 years ago

Explain the significance behind the universe having a flat geometry.

nirvana33 [79]

<span>The meaning of the Big Bang has been very often misunderstood. It is thought that something exploded somewhere and then the exploded part expanded to where we are currently. I hope this did help If not I`m so sorry.</span>

5 0

2 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago