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GaryK [48]
3 years ago
11

A 1.55 kg falcon catches a 0.415 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is

initially 27.5 m/s and the dove's velocity is 5.35 m/s in the same direction?
Physics
1 answer:
marishachu [46]3 years ago
6 0

Answer:

The answer to your question is 22.92 m/s

Explanation:

Data

Mass 1 = m1 = 1.55 kg

Mass 2 = m2 = 0.415 kg

Speed 1 = v1 = 27.5 m/s

Speed 2 = v2 = 5.35 m/s

Speed 3 = v3 = ?

Formula

   m1v1 + m2v2 = (m1 + m2)v3

- Solve for v3

   v3 = [m1v1 + m2v2] / (m1 + m2)

- Substitution

   v3 = [(1.55 x 27.5) + (0.415 x 5.35)] / (1.55 + 0.415)

- Simplification

   v3 = [ 42.63 + 2.22] / 1.965

   v3 = 44.85 / 1.965

- Result

   v3 = 22.92 m/s

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What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a space
katovenus [111]

Answer:

Part a)

Angular speed is given as

\omega = 2.50 \times 10^{-3} rad/s

Part b)

Radial acceleration of the spaceship is given as

a_c = 20.15 m/s^2

Part c)

Since we know that it is moving with constant speed

so tangential acceleration must be ZERO

Explanation:

As we know that tangential speed of the spaceship is given as

v = 29000 km/h

v = 8055.55 m/s

a)

Angular speed is given as

\omega = \frac{v}{R}

\omega = \frac{8055.55}{3220\times 10^3}

\omega = 2.50 \times 10^{-3} rad/s

Part b)

Radial acceleration of the spaceship is given as

a_c = \omega^2 R

a_c = (2.50 \times 10^{-3})^2(3220\times 10^3)

a_c = 20.15 m/s^2

Part c)

Since we know that it is moving with constant speed

so tangential acceleration must be ZERO

6 0
3 years ago
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distan
zlopas [31]

Answer:

16.6 m

Explanation:

Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.

Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.

At catch up, D = d'

d + vt = v't

d = v't - vt

d = (v' - v)t

Substituting the values of the variables into d, we have

d = (2.75 m/s - 1.17 m/s)10.5 s

d = (1.58 m/s)10.5 s

d = 16.59 m

d ≅ 16.6 m

So, the other person was 16.6 m ahead of you when you started running.

3 0
3 years ago
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