Which field in an 802.11a plcp frame are used to initialize part of the transmitter and receiver circuits?

A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w

slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0

2 years ago

1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final

tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

$\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\= \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\$

85% of energy is converted into internal energy.

$mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius$

b. The remaining 15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0

2 years ago

Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller

victus00 [196]

1.

Answer:

a) It is less

Explanation:

By energy conservation we can say that initial potential energy of both child must be equal to the final kinetic energy of the two child.

Since initially they are at same height so we will say that initial potential energy will be given as

$mgH$ and MgH

so the child with greater mass has more energy and hence smaller child will reach with smaller kinetic energy

2.

Answer:

b. The two speeds are equal.

Explanation:

As we know by mechanical energy conservation law we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

since both child starts at same height so here they both will reach the bottom at same speed

3.

Answer:

c. The two accelerations are equal

Explanation:

Since we know that average acceleration of the motion is given as

$a = \frac{v_f - v_i}{\Delta t}$

since here initial and final speeds are same so they both must have same average acceleration here.

5 0

2 years ago

Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. use each word only o

shepuryov [24]

Answer:

An annular Solar Eclipse

Explanation:

Solar eclipse is an event that occurs naturally on Earth when the moon in its orbit is positioned between the Earth and the Sun.Solar Eclipse can be total ,partial or annular.In the total solar eclipse, the moon completely covers the sun where as in the annular solar eclipse the moon covers the center of the Sun leaving outer edges of the Sun to be visible forming the<em> ring of fire.</em>In partial solar eclipse the Earth moves through the lunar penumbra as the moon moves between Earth and Sun.The moon blocks only some parts of the solar disk.Annular solar eclipse happens during new moon and the moon is at its farthest position from the Earth called Apogee.

7 0

2 years ago

The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is

Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

W = ΔEm

Where the work is

W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

Em₀ = K = ½ m v²

Final. When the system is at rest

$Em_{f}$ = $K_{e1}b +K_{e2}$ = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

∑ F = F - fr

Axis y

N- W = 0

N = W

The friction force has the equation

fr = μ N

fr = μ W

The job

W = (F – μ W) x

We substitute in the equation

(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values and solve

½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

x² 30 + 14.4 x - 1,125 = 0

x² + 0.48 x - 0.0375 = 0

We solve the second degree equation

x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

x = [-0.48 ± 0.617] / 2

x₁ = 0.0685 m

x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0

2 years ago