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faust18 [17]
3 years ago
11

Two rockets, A and B, approach the earth from opposite directions at speed 0.80c. The length of each rocket measured in its rest

frame is 100 m. What is the length of rocket A as measured by the crew of rocket B?
Physics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

The contracted length as measured is 22.22 m

Solution:

As per the question:

Speed of the rockets, v = 0.8c

Length of the rocket, L =  100 m

Now,

To calculate the length of rocket A w.r.t rocket B:

The relative speed of the rockets A and B approaching the earth from the opposite directions:

Relative speed,\ v_{r} = \frac{- 0.8c - 0.8c}{1 + \frac{v^{2}}{c^{2}}}

v_{r} = \frac{- 1.6c}{1 + \frac{(0.8c)^{2}}{c^{2}}} = 0.975c

The contracted length is given by:

L' = \frac{L}{\sqrt{1 - \frac{v_{r}^{2}}{c^{2}}}}

L' = \frac{100}{\sqrt{1 - \frac{(0.975c)^{2}}{c^{2}}}} = 22.22\ m

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otez555 [7]

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

a=3 \ m/s^2

Forces resisting motion,

F_f=432 \ N

Mass,

m = 241 \ kg

By using Newton's second law, we get

⇒ F-F_f=ma

Or,

⇒         F=ma+F_f

By putting the values, we get

⇒             =(3\times 241)+432

⇒             =723+432

⇒             =1155 \ N

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3 years ago
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diamong [38]

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Explanation:

6 0
2 years ago
A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?
katrin [286]

Answer:

<em>k = 25.18 N/m</em>

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}

If the period is known, we can find the value of the constant by solving for k:

\displaystyle k=m\left(\frac{2\pi}{T}\right)^2

Substituting the given values m=5 Kg and T=2.8 seconds:

\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2

k = 25.18 N/m

5 0
3 years ago
Fenómeno químico en virtud del cual se transforma un cuerpo o compuesto por la accion de un oxidante
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3 years ago
An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over
alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air =  1.3 kg/m³

weight of plane

From Bernoulli's principle

P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2

where 1 and 2 are two different locations at the same geo potential level  

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)

\Delta P =1227.2\ N/m^2

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

7 0
3 years ago
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