Ask question

Ask question

All categories

2 years ago

5

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the chi

ld on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.

1 answer:

Katen [24]2 years ago

8 0

Given :

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.

To Find :

If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.

Solution :

We know, for seesaw to balance :

$m_1gd_1=m_2gd_2$

Here, $d_1\ and \ d_2$ is distance from origin from pivot point.

Putting all the values in the equation, we get :

$50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet$

Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.

Hence, this is the required solution.

You might be interested in

A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

masya89 [10]

Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

$Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )$

$Q_{in} = 46.454\,kJ$

3 0

3 years ago

An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De

Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0

3 years ago

A car is about to start but it blows up. what is the problem with the car<br> ?

ratelena [41]

Answer:

because there is a bomb

6 0

3 years ago

Read 2 more answers

Carnot heat engine A operates between 20ºC and 520ºC. Carnot heat engine B operates between 20ºC and 820ºC. Which Carnot heat en

nikklg [1K]

Answer:

engine B is more efficient.

Explanation:

We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.

We also kn ow that the efficiency of Carnot cycle given as follows

$\eta =1-\dfrac{T_1}{T_2}$

Here temperature should be in Kelvin.

For engine A

$\eta =1-\dfrac{T_1}{T_2}$

$\eta =1-\dfrac{273+20}{520+273}$

$\eta =0.63$

For engine B

$\eta =1-\dfrac{T_1}{T_2}$

$\eta =1-\dfrac{273+20}{820+273}$

$\eta =0.73$

So from above we can say that engine B is more efficient.

4 0

3 years ago

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

Scilla [17]

Answer:

DIAMETER = 9.797 m

POWER = $\dot W = 28.6 kW$

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

$v =\frac{RT}{P}$

where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

$v = \frac{0.287\times 293}{100}$

v = 0.8409 m^3/ kg

from continuity equation

$A_1 V_1 = A_2 V_2$

$\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2$

$D_2 = D_1 \sqrt{\frac{V_1}{V_2}}$

$D_2 = 8 \times \sqrt{\frac{12}{8}}$

$D_2 = 9.797 m$

mass flow rate is given as

$\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}$

$\dot m = 717.309 kg/s$

the power produced $\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]$

$\dot W = 28.6 kW$

8 0

3 years ago