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3 years ago

14

You want to create a device that sends signals to your dog to stop barking. which type of electromagnetic wave would be most use

ful to investigate?

1 answer:

kati45 [8]3 years ago

6 0

The answer is radio waves. Th other waves can't send signals to a dog. Its just not how they travel.

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Why is the meteor shower is best observed after midnight?

harkovskaia [24]

Answer:

At dawn your location on earth is pointed straight in the direction of the Earth's travel in its orbit. Between midnight and dawn you are moving head-on through the location of the meteors in space, which means that you will, on average, observe more of them.

- public.nrao.edu

Explanation:

hope this helps

4 0

3 years ago

This diagram is a model of the nitrogen atom.

andrey2020 [161]

C.
Nitrogen has the atomic no. of 7 so the proton and the electron no. is 7.
Three more electrons are needed to fulfil the octet structure: 2,8

7 0

2 years ago

which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla

Yakvenalex [24]

Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean

7 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

5 0

3 years ago