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3 years ago

What are various paradox related to time travel?

Physics

1 answer:

erma4kov [3.2K]3 years ago

5 0

Answer:

This is a paradox — an inconsistency that often leads people to think that time travel cannot occur in our universe." A variation is known as the "grandfather paradox" — in which a time traveler kills their own grandfather, in the process preventing the time traveler's birth

Explanation:

hope this Wil help ....

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An airplane, flying horizontally at an altitude of 3 miles, passes directly over an observer. If the constant speed of the air

natima [27]

I’m not sure if this helped you so here you go

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3 years ago

Anyone know these questions?

salantis [7]

400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent

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3 years ago

A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the l

Virty [35]

Answer:

N = 337.96 N

Explanation:

∅ = 32º

F = 249 N

m = 21 Kg

N = ?

We can apply:

∑ F = 0 (↑)

- Fy - W + N = 0 ⇒ N = Fy + W

⇒ F*Sin ∅ + m*g = N

⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)

⇒ N = 337.96 N (↑)

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3 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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3 years ago

A 0.2 kg cannon ball is fired at an upward angle of 45° from the top of a 165 m cliff with a speed of 175 m/s. (A) Using conserv

sergejj [24]

To solve this problem, it is necessary to apply the concepts related to the work done by a body when a certain distance is displaced and the conservation of energy when it is consumed in kinetic and potential energy mode in the final and initial state. The energy conservation equation is given by:

$\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f$