Ask question

Ask question

All categories

iogann1982 [59]

2 years ago

9

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

62-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects. The positive x-direction is to the right, and the origin is at the center of the chair.

1 answer:

Stells [14]2 years ago

7 0

Answer:

Komila should sit 0.33m from the middle of the board towards tahreen.

Explanation:

We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.

From the diagram,

F_d is force exerted by dan

F_t is force exerted by tahreen

F_k is force exerted by komila

F_b is force of board at the mid point.

x1 is distance of dan from the centre of the chair

x2 is distance of komila from the centre of the chair

x3 is distance of tahreen from komila

We are given;

Mass of Dan;m_d = 62 kg

Mass of tahreen;m_t = 50 kg

Mass of komila;m_k = 54 kg

Now, taking moments about the centre of the chair, we have;

(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0

Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g

We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;

x1 = (x2 + x3) = 1.5

Thus, we now have;

(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0

F_d = m_d*g = 62 * 9.8 = 607.6 N

F_t = m_t * g = 50 x 9.8 = 490 N

F_k = m_k * g = 54 x 9.8 = 529.2 N

So plugging in these values, we have;

(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0

911.4 - 735 = 529.2 x2

529.2 x2 = 176.4

x2 = 176.4/529.2

x2 = 0.33m

Komila should sit 0.24m from the middle of the board towards tahreen

You might be interested in

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago

Traffic officials indicate, it takes longer to ______ when you drive fast.

nignag [31]

The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.

7 0

2 years ago

Read 2 more answers

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago

Air at 27 ºC and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligibl

Citrus2011 [14]

Answer:

27 °C

Explanation:

BY the statement of the question it is clear that it is about an ideal gas - and hence if change in KE is about zero - then there will be no change of temperature.

So, answer is 27 °C

4 0

2 years ago

You are target shooting using a toy gun that fires a small ball at a speed of 15 m/s. When the gun is fired at an angle of 31 de

Ray Of Light [21]

Answer: 13.1° from the horizontal

Explanation: For projectile

Horizontal distance R of a projectile is:

R= U×Usin2x/g

Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s

R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m

Therefore if R is halved.

R/2 = 20.27/2=10.14m

Hence the angle for this case is.

Making sin(2x) the subject of formula

Sin2x= Rg/U×U

R is now 10.14 in this case.

Sin2x= 10.14×9.8/15×15=0.4415

2x=arcsin0.4415=26.2

x= 26.2/2

x= 13.1°

Good luck...

3 0

2 years ago