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3 years ago

How do repeater stations improve the quality of a broadcast?

1 answer:

3 0

They pick up broadcast signals, amplify them, then send them out.
This is needed due to the radio waves traveling in a ripple pattern; they degrade over time.

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2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA

11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *144

M_c = 2.5*( 42 / 150 ) *144

M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0

3 years ago

Indicate the result of each of the following unit vector cross products (unit vector hat-symbols not shown): . kxi = . jxi= -jxk

notka56 [123]

Answer:

Explanation:

The cross product of two vectors is given by

$\overrightarrow{A}\times \overrightarrow{B}=\left | \overrightarrow{A} \right |\left | \overrightarrow{B} \right |Sin\theta \widehat{n}$

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.

Here, K x i = - j

As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.

The direction of cross product of two vectors is given by the right hand palm rule.

So, k x i = j

j x i = - k

- j x k = - i

i x i = 0

4 0

2 years ago

Please Its urgent I need your HELP!!!!!!!!!!!!!!!!!!!!!!!

ozzi

Raising the temperature results in the radiator giving off photons of high-energy ultraviolet light. As heat is added, the radiator emits photons across a wide range of visible-light frequencies

4 0

2 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.

LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

The moment of inertia of the system is 289.088 kg.m^2

7 0

3 years ago

A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu

Free_Kalibri [48]

Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) = 2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

3 0

2 years ago

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