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finlep [7]
3 years ago
9

Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down t

he runway. A smaller plane with the same acceleration has a takeoff speed of 28.2 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?
Physics
1 answer:
V125BC [204]3 years ago
6 0

Answer:

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

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One event occurs at the origin at t equal to zero, and a second events occurs at the point x=5m along the x-axis at time with ct
Anastasy [175]

Answer:

  • The separation will be spacelike.
  • The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

Explanation:

We can define the separation between two events (using the + - - - signature)  as :

(\Delta s )^2  = (ct_2 - c t_1 )^2 - (x_2 - x_1)^2

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.

For our problem

c t_1 = 0

x_1 = 0

ct_2 = 4 \ m

x_2 = 5 \ m

(\Delta s )^2  = (4 \ m - 0 )^2 - ( 5 \ m - 0)^2

(\Delta s )^2  = (4 \ m )^2 - ( 5 \ m 0)^2

(\Delta s )^2  = 16 \ m^ 2 - 25 \ m^2

(\Delta s )^2  = - 9\ m^2

So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

8 0
3 years ago
What is a conductor
Soloha48 [4]

C. none of the above

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8 0
3 years ago
Read 2 more answers
Australia has 3 a’s but all pronounced differently how?
kobusy [5.1K]

BecausE the first 'a' is used with a 'u' making a "ahh" sound

The second 'a' isn't paired with any other vowels so it's sound is a strong A sound

And the 'a' at the end is paired with an 'i' in front of it making an "ee-uhh" sound


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8 0
3 years ago
A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th
finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

\omega _{0}=\frac{1}{\sqrt{LC}}

\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = \sqrt{R^{2}+\left ( XL - Xc \right )^{2}}

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

7 0
3 years ago
What is the temperature of a star (in Kelvin) if its peak wavelength is 150 nm (that is, 150 x 10^-9 m)?
liq [111]

Answer:

19320 K

Explanation:

The temperature of a star is related to its peak wavelength by Wien's displacement law:

T=\frac{b}{\lambda}

where

T is the absolute temperature at the star's surface

b=2.898 \cdot 10^{-3}m\cdot K is Wien's displacement constant

\lambda is the peak wavelength

Here we have

\lambda=150 nm = 150\cdot 10^{-9}nm

Substituting into the equation, we find

T=\frac{2.898\cdot 10^{-3}}{150\cdot 10^{-9}}=19320 K

7 0
4 years ago
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