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4 years ago

A 175-kg motorcycle goes around an unbanked turn of radius 75.4 m at a steady 112 km/h. Determine its acceleration?

Physics

1 answer:

Nuetrik [128]4 years ago

7 0

Answer: $12.83 m/s^{2}$

Explanation:

In this case we are talking about the centripetal acceleration $a_{c}$ , which can be calculated by:

$a_{c}=\frac{V^{2}}{r}$

Where:

$V=112 \frac{km}{h} \frac{1000 m}{1 km} \frac{1h}{3600 s}=31.11 m/s$ is the speed of the motorcycle

$r=75.4 m$ is the radius of the path the motorcycle is going around

Solving:

$a_{c}=\frac{(31.11 m.s)^{2}}{75.4 m}$

Finally:

$a_{c}=12.83 m/s^{2}$

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Vaselesa [24]

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5 0

4 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is: