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3 years ago

9

5

2 answers:

MakcuM [25]3 years ago

8 0

Answer:

I believe the answer is triangle

Hope it helps!

lana [24]3 years ago

4 0

C because your cutting from the side the the pointy part would make a triangle

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For an individual ( not a repair station) to conduct a complete 100 hour inspection on an aircraft and approve it for return to

Oliga [24]

Answer:

For an individual ( not a repair station) to conduct a complete 100 hour inspection on an aircraft and approve it for return to service requires a mechanic certificate with and powerplant and airframe ratings

Explanation:

Because these are important certificates issued by the FAA giving the mechanic authority to inspect to inspect an aircraft and approve its return to services

8 0

3 years ago

A 1.13 kg skateboard is coasting along the pavement at a speed of 4.28 m/s when a 0.93

faust18 [17]

By conservation of momentum,

Pinitial = Pfinal

m1v1 + m2v2 = (m1 + m2)*vf

m1 = mass of skateboard = 1.13 kg

m2 = mass of cat = 0.93 kg

v1 = initial velocity of skateboard = 4.28 m/s

v2 = initial velocity of cat = 0 m/s

vf = final velocity of skateboard-cat combo

So plug in the values and solve for vf,

1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf

vf = 2.35 m/s

5 0

3 years ago

An electron with an initial speed of 660,000 m/s is brought to rest by an electric field. What was the initial kinetic energy of

GREYUIT [131]

Answer:

Kinetic energy of the electron $E_k=1.23\ eV$

Explanation:

It is given that,

Initial speed of an electron, u = 660000 m/s

Final speed of the electron, v = 0 (at rest)

The kinetic energy of an electron is possessed due to the motion of an electron. The mathematical formula for the kinetic energy is given by :

$E_k=\dfrac{1}{2}mv^2$

m is the mass of electron

$E_k=\dfrac{1}{2}\times 9.1\times 10^{-31}\times (660000)^2$

$E_k=1.98\times 10^{-19}\ J$

Since, $1\ eV=1.602\times 10^{-19}\ J$

$E_k=\dfrac{1.98\times 10^{-19}}{1.602\times 10^{-19}}$

$E_k=1.23\ eV$

So, the initial kinetic energy of the electron is 1.23 eV. Hence, this is the required solution.

7 0

4 years ago

The diagrams show objects’ gravitational pull toward each other. 3 diagrams labeled X, Y, and Z. X: 2 circles approximatley 2 in

Dafna11 [192]

Answer:C

Explanation:

6 0

4 years ago

Read 2 more answers

When a certain gas under a pressure of 4.90 106 pa at 20.0°c is allowed to expand to 3.00 times its original volume, its final p

Elina [12.6K]

As per question the initial states of the gases are given as

INITIAL STATE: FINAL STATE:

$p_{1} =4.90106 pa$ $p_{2} =1.06106 pa$

$v_{1} =v[say]$ $v_{2} =3v[say]$

$T_{1} =20 degree celcius =293 K$ $T_{2} =?$

AS per combined gas equation obtained from the combination of Boyle's law and Charles law [Basic ideal gas laws]

$\frac{p_{1} v_{1} }{T_{1} } =\frac{p_{2}v_{2} }{T_{2} }$

Hence $T_{2} =\frac{p_{2} v_{2}T_{1} }{p_{1} v_{1} }$

= $\frac{1.06106*3v*293}{4.90106*v}$

=190.3 K [ANS]

6 0

3 years ago