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grin007 [14]
3 years ago
10

What is the acceleration of a skier that goes from 2.50 m/s to 14.5 m/s while traveling 505 m down a slope?

Physics
1 answer:
Alik [6]3 years ago
3 0

Answer:

493 m*m/s

Explanation:

14.5-2.50=12

505 x 12=493

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A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2
Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass m_1 = 10 g= 10^{-2} \ kg\\\\

initial speed u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\

block mass m_2 = 4\ Kg

initial speed v_2 =-4.2 \frac{m}{s}

final speed v_2= 0

Let v_1 will be the bullet speed after collision:

throughout the consevation the linear moemuntum

\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\

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The kinetic energy of the bullet in its emerges from the block

k=\frac{1}{2} m_1 v_1^2

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3 0
3 years ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

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(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

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

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

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Answer:

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And depending on direction the greater force is being pulled towards
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