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il63 [147K]
3 years ago
13

A piece of aluminium foil is held between a bar magnet and a paperclip. What will happen to the paper clip?

Physics
1 answer:
Nuetrik [128]3 years ago
7 0

Answer:the paper clip will be attracted to the magnet

Explanation:

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Sorry!

This cannot be answered. We don't have weight, height, etc.

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A child’s toy arrow is launched horizontally from a bow at a speed of 8 m/s from a second story window 4 meters above the ground
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First find the time it takes to hit the sand t
using s=1/2 g t^2
t^2= 8/9.8= 0.82 seconds
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The horizontal speed stays constant at 8 m/s
The vertical speed =.82 * 9.8 = 8.0 m/s
As the two velocities are equal the angle must be 45 degrees to the horizontal
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Why are bulb filaments made of metals and not wood
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If the torque required to loosen a nut that
liraira [26]

Explanation:

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7 0
3 years ago
2. In the same tournament, a player is positioned 35 m (40° W of S] of the net. He shoots the puck
andrezito [222]

Answer:

The displacement of the net from player 2 in component form = (-47.498î - 26.812j)

The displacement of the net from player 2 in statement form is 54.54 m and 29.44° (S of W) or 60.56° (W of S)

Explanation:

The sketch of the bearings described in the question is presented in the attached image to this solution.

Method 1

Using component method

Taking the player 1's position as the origin,

The displacement of the player 2 from the origin is (25î) m

The displacement of the net from the origin is 35[(sin θ)î + (cos θ)j]

But θ is the angle of the net's displacement reading from the positive x-axis in the anticlockwise direction. θ = 230°

Displacement of the net from the origin = 35[(cos 230°) + (cos 230°)]

= 35[-0.6428î - 0.7660j]

= (-22.498î - 26.812j) m

In component form, taking note of the directions of the respective displacements calculated (check the attached image)

(The displacement of the net from player 1) = (The displacement of player 2 from player 1) + (The displacement of the net from player 2)

Since we have agreed that player 1 is the origin

(The displacement of the net from origin) = (The displacement of player 2 from origin) + (The displacement of the net from player 2)

(-22.498î - 26.812j) = (25î) + (The displacement of the net from player 2)

The displacement of the net from player 2 = (-22.498î - 26.812j) - (25î) = (-47.498î - 26.812j)

The magnitude of this displacement = √[(-47.498)² + (-26.812)²]

= √(2,256.060004 + 718.883344) = 54.54 m

Direction = tan⁻¹ (-26.812/-47.498) = 209.44° (the signs on the components show that the direction is the third quadrant from the positive x-axis in the anti-clockwise direction)

Hence, the displacement of the net from player 2 is 54.54 m and 29.44° (S of W)

Method 2

Using trignometry,

We will use cosine and sine rule to obtain the required magnitude and direction of the displacement of the net from player 2

Cosine rule

Magnitude = √[35² + 25² - (2×25×35×cos 130°)] = √2,974.8783169514 = 54.54 m

Sine rule

(Sin θ)/35 = (Sin 130°)/54.54

Sin θ = (35 × Sin 130°)/54.54 = 0.4916

θ = Sin⁻¹ (0.4916) = 29.44°

This answer matches the answers from method 1.

Hope this Helps!!!

5 0
3 years ago
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