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Minchanka [31]
2 years ago
14

Skater N, 58.9 kg, is moving north at a speed of 7.8 m/s when she collides with Skater E, 72.6 kg, moving east at 3.5 m/s. The t

wo skaters are stuck together. In what direction do they move after the collision?
a. 61 degrees c. 32 degrees

b. 45 degrees d. 27 degrees
Physics
1 answer:
Alik [6]2 years ago
4 0

The direction of the skaters after the collision is 61 degrees.

<h3>Momentum of the skaters</h3>

The momentum of the skaters is calculated as follows;

P(N) = 58.9 x 7.8 = 459.42 kgm/s

P(E) = 72.6 x 3.5 = 254.1 kgm/s

<h3>Direction of the skaters after collision</h3>

The direction of the skaters after the collision is calculated as follows;

\theta = tan^{-1} (\frac{459.42}{254.1} )\\\\\theta = 61^ 0

Thus, the direction of the skaters after the collision is 61 degrees.

Learn more about direction of collision here: brainly.com/question/25891473

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Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

v=\frac{d}{t}

where

d is the displacement

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In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

v=\frac{15 km}{2.0 h}=7.5 km/h

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

v=\frac{15,000 m}{7200 s}=2.1 m/s

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The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
Setler79 [48]

Answer:

0.72

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T = Time period of oscillation = 1.5 s

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w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s

A = Amplitude of oscillation = 40 cm = 0.40 m

\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

a = Aw^{2}

frictional force is given as

f = \mu mg

As per newton's second law

f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

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Answer:

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