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Sliva [168]
3 years ago
6

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expa

nding when it is heated from 10.9 °C to 40.0 °C?
Physics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water (B), in newtons per square meters, can be estimated by means of the following model:

B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}} (1)

Where:

\rho_{o} - Water density at 10.9 °C, in kilograms per cubic meter.

\rho_{f} - Water density at 40 °C, in kilograms per cubic meter.

\Delta P - Pressure change, in pascals.

If we know that \rho_{o} = 999.623\,\frac{kg}{m^{3}}, \rho_{f} = 992.219\,\frac{kg}{m^{3}} and B = 2.2\times 10^{9}\,\frac{N}{m^{2}}, then the bulk modulus of water is:

\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)

\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)

\Delta P = -16294943.19\,Pa \,(-160.819\,atm)

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

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A 10 kilogram sled is pulled across a frictionless surface with a force of 50 newtons for a distance of 10 meters. The pull is a
Gala2k [10]

Answer:

The power will be "250 watt". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 50 N

Displacement,

d = 20 m

Time,

t = 2.0 seconds

Whenever the block is pulled, the angle will be "0" i.e., Cos0° = 1

Now,

The work will be:

= Force\times displacement\times \Theta

On substituting the given values, we get

= 50\times 10\times Cos0^{\circ}

= 50\times 10\times 1

= 500 \ Newton

Now,

The Power will be:

= \frac{Work \ done}{time}

= \frac{500}{2.0}

= 250 \ watt

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Momentum
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Answer:

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3 years ago
When a substance is poured from one container to another, its shape changes but its volume does not, what state of matter is the
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Answer: When a liquid substance is poured into a vessel, it takes the shape of the vessel, and, as long as the substance stays in the liquid state, it will remain inside the vessel. Furthermore, when a liquid is poured from one vessel to another, it retains its volume (as long as there is no vaporization or change in temperature) but not its shape.

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8 0
3 years ago
The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o
WINSTONCH [101]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which \mu k = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant \theta = 30^{o}.

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           \sum F_{y} = 0

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        \sum F_{x} = 0

       T_{2} - F_{k} = 0

         T_{2} = \mu_{k} N

                  = 0.4 \times 75 lb

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         \sum F_{x} = 0

       T Cos (30^{o}) + T Cos (30^{o}) = T_{2}

          2T Cos (30^{o}) = T_{2}

    or,             T = \frac{T_{2}}{2 Cos (30^{o})}

                        = \frac{30}{2 Cos (30^{o})}

                        = \frac{30}{1.732}

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = 17.32 lb \times 6

                = 103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

3 0
3 years ago
Read 2 more answers
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