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Kitty [74]
3 years ago
8

A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the roc

ky beach 2.10 s later. just before the clam lands, what is (a) its horizontal component of velocity, and (b) its vertical component of velocity? (c) how would your answers to parts (a) and (b) change if the speed of the crow were increased? explain.
Physics
1 answer:
vovangra [49]3 years ago
5 0

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

v_x = 2.70 m/s

Part b)

In vertical direction we can use kinematics

v_f = v_i + at

v_f = 0 + 2.1 * 9.8

v_f = 20.6 m/s

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

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The resultant of two forces is 250 N and the same are inclined at 30° and 45° with resultant one on either side calculate the ma
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Answer:

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Explanation:

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The (magnitude) of the resultant of two forces = 250 N

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Let, F₁ and F₂ represent the two forces, we have;

F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force

The components of F₁ are \underset{F_1}{\rightarrow} = -F₁ × sin(30°)·i + F₁ × cos(30°)·j

The components of F₂ are \underset{F_2}{\rightarrow} = F₂ × sin(45°)·i + F₂ × cos(45°)·j

The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j

The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out

We have;

F₂ × sin(45°)·i = F₁ × sin(30°)·i

F₂ ·√2/2 = F₁/2

∴ F₁ = F₂ ·√2

∴ F₂ × cos(45°)·j  + F₁ × cos(30°)·j = 250·j

Which gives;

F₂ × cos(45°)·j  + F₂ ·√2 × cos(30°)·j = 250·j

F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ × ((√2)/2 × (1 + √3))·j = 250·j

F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N

F₂ ≈ 129.41 N

F₁  = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N

F₁  ≈ 183.013 N

The two forces are;

A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.

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