Ask question

Ask question

All categories

4 years ago

14

A 45.0 kg girl is standing on a 159 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.38 m/s relative to the plank.(a) What is her velocity relative to the surface of ice?1 m/s(b) What is the velocity of the plank relative to the surface of ice?2 m/s

1 answer:

Lady_Fox [76]4 years ago

8 0

Answer:

Explanation:

Given

Mass of girl $m=45 kg$

Mass of Plank $M=159 kg$

let $v_p$ and $v_g$ be the velocity of Plank and girl w.r.t ice

therefore $v_{gp}=1.38 m/s$

Conserving Momentum

$(M+m)v_p+mv_{gp}=0$

$(45+159)v_p=-45\cdot v_{gp}$

$v_p=-\frac{45}{204}\cdot 1.38$

$v_p=-0.304 m/s$

$v_{gp}=v_g-v_p=1.38$

$v_g=-0.304+1.38=1.075 m/s$

You might be interested in

When discussing Newton's laws of motion,which terms do people most likely yse when talking about Newton's third law of motion?

mestny [16]

When discussing Newton's laws of motion, particularly Newton's third law of motion, the terms that almost everyone will use are "action" and "reaction".

You must not take this to mean that they understand what they're talking about.

4 0

3 years ago

Need help with these please

Montano1993 [528]

Answer:

Explanation:

Which number all or one?

6 0

3 years ago

Read 2 more answers

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

a)

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

$h_m=0.01838\ m=1.838\ cm$

<u>Now the difference in the column is :</u>

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

7 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

A wave of water moving up a river, initiated by tidal action and normal resonances within a river estuary, is called a:

IgorC [24]

Answer:

friction solar system

Explanation:

6 0

3 years ago