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4 years ago

What are the three pins that may be worn below your name on the ffa jacket?

1 answer:

8 0

<span>Highest degree, highest office and highest award are the three pins that may be worn below your name on the ffa jacket.
</span>FFA is the biggest of the profession and specialized student associations in U.S. schools.The National FFA Organization is an adolescent initiative association that has a constructive outcome in the lives of youngsters by building up their potential for head authority, self-awareness and profession accomplishment through education of agriculture.The Motto of FFA is;
"Learning to do, Doing to learn, Earning to live, Living to serve."

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If anyone has course hero could you send me the Gizmo Fan Cart Physics?

azamat

Answer:

I might be able to help, but I also might not. I need to check if I can get you the answer.

Explanation:

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3 years ago

When an object is moving with uniform circular motion, the centripetal acceleration of the object a. is circular. b. is perpendi

never [62]

Hi Pupil Here's Your answer :::

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When an object is moving with uniform circular motion, the centripetal acceleration of the object $\textbf{is\:Zero}$ .

Because the body is moving with a uniform velocity and hence, there might not be any acceleration due to it.

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Hope this helps .......

5 0

3 years ago

Satelite communication is only possible using

zimovet [89]

None of the choices is correct.

Satellite communication is quite effective using either analog OR digital transmission.

Sound can't be used for satellite communication at all. (There's no medium to carry the sound to or from the satellite, and if there were, then the round-trip up and down through a geostationary satellite would take at least a minute at the sea-level speed of sound.)

Cellular devices can work but aren't required.

6 0

3 years ago

Read 2 more answers

Which of the following is a risk associated with the application of biotechnology?

dmitriy555 [2]

Answer:

d.) potential for the exchange of genetic material

Explanation:

3 0

3 years ago

A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what spee

GenaCL600 [577]

We will have the following:

First, the equation to use is the following:

$d=v_ot+\frac{1}{2}at^2$

Now, we transform the total distance the cat would need to travel:

$90\operatorname{cm}+60\operatorname{cm}=150\operatorname{cm}\cdot\frac{1m}{100\operatorname{cm}}=1.5m$

So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

$t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots$

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

$1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)$ $\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots$

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.

5 0

2 years ago