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2 years ago

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While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

not completed before crossing the road or reference line. This would most likely occur in turn:_______

1 answer:

erica [24]2 years ago

5 0

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

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A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

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2 years ago

Use the formula a=h/n to find the batting average a of a batter who has h hits in n times at bat. Solve the formula for h. if a

Inessa [10]

A = h / n => h = a*n

a = 0.290 hit / time
n = 300 times

=> h = 0.290 hit / time * 300 time = 87 hits

Answer: 87 hits

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2 years ago

A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen

saw5 [17]

Answer:

The electric field strength is $4.5\times 10^{4} N/C$

Solution:

As per the question:

Area of the electrode, $A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}$

Charge, q = 50 nC = $50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m$

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

$\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}$

Now, the electric field strength of the electrode is:

$\vec{E} = \frac{\sigma}{2\epsilon_{o}}$

where

$\epsilon_{o} = 8.85\times 10^{- 12} F/m$

$\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}$

$\vec{E} = 4.5\times 10^{4} N/C$

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2 years ago

A ray of light is traveling from water into a diamond. Upon hitting the diamond, a light ray bends toward the normal. With this

avanturin [10]

The correct answer is b

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2 years ago

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A particular metal has a work function of 1.05 eV. A light is shined onto this metal with a corresponding wavelength of 324 nm.

LenaWriter [7]

Answer:

Velocity of electron will be $v=0.986\times 10^6m/sec$

Explanation:

We have given work function of metal $\Phi =1.05eV=1.05\times 1.6\times 10^{-19}J=1.68\times 10^{-19}J$

Wavelength of the light $\lambda =324nm=324\times 10^{-9}m$

So energy is given by $E=\frac{hc}{\lambda }$ , here h is plank's constant and c is speed of light

So $E=\frac{6.6\times 10^{-34}\times3\times 10^8}{324\times 10^{-9} }=6.11\times 10^{-19}j$

For a metal we know that $E=\Phi +KE_{MAX}$

So $KE_{MAX}=E-\Phi =6.11\times 10^{-19}-1.68\times 10^{-19}=4.43\times 10^{-19}$

Now kinetic energy is given by $KE=\frac{1}{2}mv^2$

$4.43\times 10^{-19}=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=0.986\times 10^6m/sec$

So velocity of electron will be $v=0.986\times 10^6m/sec$

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2 years ago