Ask question

Ask question

All categories

LuckyWell [14K]

2 years ago

13

After a 60-kg person steps from a boat onto the shore, the boat moves away with a speed of 0.60 m/s with respect to the shore. I

f the same person steps from the shore onto the boat, the person and the boat move away from the shore with the speed of 0.40 m/s. Assume that the initial speed of the person with respect to the shore is the same in both cases and that the person is moving in a horizontal direction. Determine the Mass of the boat and the speed of the person.

1 answer:

9966 [12]2 years ago

7 0

Answer:m=80 kg

Explanation:

Given

mass of Person $M=60 kg$

When Person Steps From boat onto shore boat moves away to 0.6 m/s away from shore

let v be the velocity of Person and m be the mass of boat

Conserving momentum

$M v-0.7 m=0$

$M v-0.7 m=0$

$60 v=0.7 m$ ----1

When Person step on to boat, boat moves on to left 0.4 m/s towards left

$Mv=(M+m)\cdot 0.4$

$60 \cdot v=(60+m)\cdot 0.4$ ---------2

using 1 and 2

$60 v=(60+85.71 v)\cdot 0.4$

$v=0.933 m/s$

therefore $m=79.99\approx 80 kg$

You might be interested in

Four students were loading boxes of food collected during a food drive. The force that each student exerted while lifting and th

Fudgin [204]

Answer:

it is B

Explanation:

Edge2020

7 0

2 years ago

Read 2 more answers

calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it

Lera25 [3.4K]

The force of F=10 N produces an extension of
$x=4 cm=0.04 m$
on the string, so the spring constant is equal to
$k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m$

Then the string is stretched by $\Delta x=40 cm=0.40 m$ . The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
$W= \Delta U= \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J$

5 0

2 years ago

Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?

Svetradugi [14.3K]

Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>

6 0

2 years ago

Read 2 more answers

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

2 years ago

Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stu

Dvinal [7]

Answer:

The current is $I = 1.1434*10^{-5}}\ A$

Explanation:

From the question we are told that

The radius of the kite string is $R = 2.02 mm = 0.00202 \ m$

The distance it extended upward is $D = 0.823 km = 823 \ m$

The thickness of the water layer is $d = 0.506 mm = 0.000506 \ m$

The resistivity is $\rho = 159\ \Omega \cdot m$

The potential difference is $V = 186 MV = 186 *10^{6} \ V$

Generally the cross sectional area of the water layer is mathematically represented as

$A = \pi r^2$

Here r is mathematically represented as

$r = [(R + d ) - R]$

=> $r = [(0.00202 + 0.000506 ) - 0.00202]$

=> $r = 0.000506$

=> $A = 3.142 * [0.000506]^2$

=> $A = 8.0447*10^{-7}\ m^2$

Generally the resistance of the water is mathematically represented as

$R = \frac{\rho * D }{A}$

=> $R = \frac{159 *823 }{8.0447*10^{-7}}$

=> $R = 1.62662 * 10^{11} \ \Omega$

Generally the current is mathematically represented as

$I = \frac{V}{R}$

=> $I = \frac{186 *10^{6} }{1.62662 * 10^{11}}$

=> $I = 1.1434*10^{-5}}\ A$

8 0

2 years ago