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r-ruslan [8.4K]
3 years ago
8

A light ray moving at a 54.3 deg

Physics
1 answer:
MrMuchimi3 years ago
3 0

Answer:

45.25^{\circ}

Explanation:

Applying Snell's law

n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}

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Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N

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Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm
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3 years ago
two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h
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Answer:

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Explanation:

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3 years ago
A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
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Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

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We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

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The initial speed of the tiger is 1.80 m/s

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