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3 years ago

What airplanes did the ffa ground earlier this year after several crashes?

Physics

1 answer:

frez [133]3 years ago

8 0

The FFA is the Future Farmers of America. It has no authority in aeronautical matters.

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A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.

nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

$v = r\omega \rightarrow \omega = \frac{v}{r}$

Here,

v = Lineal velocity

$\omega$ = Angular velocity

r = Radius

Our values are

$v = 6/ms$

$r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m$

Replacing to find the angular velocity we have,

$\omega = \frac{6m/s}{0.06m}$

$\omega = 100rad/s$

Convert the units to RPM we have that

$\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})$

$\omega = 955.41rpm$

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

4 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers

What happenes to our body temperature on a cold winter day or a hot summer day

viva [34]

In cold winter day, the body temperature falls down from normal temperature of 98.6°F (37°C) to 95°F (35°C). In winter body losses heat faster than it generates heat. If the temperature fall further below 95°F (35°C), it is emergency condition known as Hypothermia. One has to consult doctor in this case.

In summer hot days, body evaporates water in the form of sweat, in order to remain itself cool. Rise of temperature up to 100°F is normal. It is recommended to hydrate body to maintain temperature in summer days.

3 0

3 years ago

As you can see, my cousin has a lot of hair. He uses an 1800 W blow dryer and it takes him

maw [93]

Power = 1800W (or 1.8KW by dividing by 1000)

Time = 3 hours

Power = energy/ time

1.8KW = energy/ 3

5.4Kw/h= energy

(5.4KJ or 5400J used)

$0.15 Kw/h

$0.15 X 5.4 = 0.81

Thus, cost $0.81

Hope this helps!

5 0

2 years ago