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2 years ago

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A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an obser

ver on Tatooine, the cruiser is traveling away from the planet with a speed of 0.6c. The pursuit ship is traveling at a speed of 0.8c relative to Tatooine, in the same direction as the cruiser. What is the speed of the pursuit ship relative to the cruiser?

1 answer:

Ede4ka [16]2 years ago

4 0

Answer:

0.384c

Explanation:

To find the speed of the pursuit ship relative to the cruiser you use the following relativistic equation:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

u': relative speed

u: speed of the pursuit ship = 0.8c

v: speed of the cruiser = 0.6c

c: speed of light

You replace the values of the parameters to obtain u':

$u'=\frac{0.8c-0.6c}{1-\frac{(0.6c)(0.8c)}{c^2}}=0.384c$

Hence, the relative speed is 0.384c

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On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som

k0ka [10]

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

$v=f \lambda$

Where:

$v$ is the speed

$f=\frac{1}{T}$ is the frequency, which has an inverse relation with the period $T=1 h$

$\lambda=800 km$ is the wavelength

Solving with the given units:

$v=\frac{1}{T}\lambda$

$v=\frac{1}{1 h}800 km$

$v=800.00 km/h$ This is the speed of the wave in km/h

Transforming this speed to m/s:

$v=800.00 \frac{km}{h} \frac{1 h}{3600 s} \frac{1000 m}{1 km}$

$v=222.22 m/s$ This is the speed of the wave in m/s

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3 years ago

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the di

Sidana [21]

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$ (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$

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3 years ago

A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric

vichka [17]

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

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Answer:

it's C endothermic

Explanation:

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Answer:

The answer is 29 AU

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