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gregori [183]
3 years ago
6

A 300 g bird is flying along at 6.0 m/s and sees a 10 g insect heading straight towards it with a speed of 30 m/s. The bird open

s its mouth wide and swallows the insect. a. What is the birds speed immediately after swallowing the insect? b. What is the impulse on the bird? c. If the impact lasts 0.015 s, what is the force between the bird and the insect?
Physics
1 answer:
Gelneren [198K]3 years ago
5 0

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s  (b)

F*t= I

F= I/t

F= 20 N (c)

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An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
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Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

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A 1750kg bumpercar moving at 1.50m/s to the right collides elastically with a 1450kg car going to the left at 1.10m/s. The 1750k
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An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1
Ksivusya [100]
<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

                 F=\frac{GMm}{r^2}

           Where G =  6.67 x 10⁻¹¹ N m²/kg²

                       M = Mass of body 1

                       M = Mass of body 2

                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

                 m = Mass of asteroid = 4.00×10¹⁶ kg

                 F = 3.14×10¹³ N

Substituting

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The asteroid is 4.11 x 10¹¹ m far from Sun

3 0
3 years ago
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