Physics: In an attempt to reduce the extraordinary long travel times for voyaging to distant stars, some people have suggested traveling close to the speed of light. suppose you wish to visit the red giant star betelgeuse, which is 430 ly away, and that you want your 20,000kg rocket to move so fast that you age only 20 years during the round trip. how fast must the rocket travel relative to earth?

In an attempt to reduce the extraordinary long

Ответ:

Redhead667

17.01.2022

Physics

0.99973 c

Explanation:

Betelgeuse is 430 ly away from earth = 4.07 ×10¹⁸ m

You want to age only 20 years during the round trip. It means you want to cover one way distance in 10 years relative to Earth.

10 years = 3.15 × 10⁸ s

This is possible only when rocket travels close to speed of the light.

We will use time dilation formula:

$t_o=t' \sqrt {1-\frac{v^2}{c^2}}$

where, t_o is the time elapsed in Earth's frame of reference, t' is the time elapsed in rocket's frame of reference, v is the velocity of the rocket and c is the speed of light.

$t_o=\frac{d}{v} \sqrt {1-\frac{v^2}{c^2}}$

where d is the distance of Betelgeuse from Earth.

$\Rightarrow t_o\times v=d \sqrt {1-\frac{v^2}{c^2}}$

$\Rightarrow t_o^2v^2=d^2-\frac{d^2v^2}{c^2}$

$(t_o^2+\frac{d^2}{c^2})v^2=d^2\\ \Rightarrow v=\frac{d}{\sqrt {t_o^2+\frac{d^2}{c^2}}}$

$\Rightarrow v = \frac{4.07\times10^{18} m}{\sqrt{(3.15\times10^8 s)^2+\frac{(4.07\times 10^{18}m)^2}{(3\times 10^8m/s)^2}}}=\frac{4.07\times 10^{18} m}{1.36\times 10^{10}s}=2.99\times10^8m/s = 0.99973 c$

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Ответ:

uaine2002

08.02.2022

Physics

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

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