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4 years ago

If I move 15ft foward, 15 ft backwards, 15 ft to the right, 15ft to the left where am I?

Physics

2 answers:

Naya [18.7K]4 years ago

8 0

Exactly where you started

lisov135 [29]4 years ago

4 0

-- When you finish those moves, you are exactly where you started from.

-- Your distance covered is 60 ft.

-- Your displacement is zero.

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Which best describes what happens when light traveling through air enters water at an angle?

vovangra [49]

I think it is "A'' 'It moves along straight lines in air and changes direction when it enters water.

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4 years ago

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The rate at which speed changes is called

sergiy2304 [10]

Answer:

meters per second

Explanation:

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3 years ago

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Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f

o-na [289]

Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m

μo = 4 π × 10^-7 Tm/A

The formula for the force per unit length between the two wires is given by

F = μo / 4π x (2 i1 x i2) / d

3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082

I = 0.0362 A = 36.22 mA

4 0

4 years ago

Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir

tia_tia [17]

Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$

The Energy lost in the collision is 7560 Joules

7 0

3 years ago

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

$A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})$

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

$A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )$

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

7 0

3 years ago