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liberstina [14]
3 years ago
8

Anyone know how to do this?

Physics
1 answer:
Nadya [2.5K]3 years ago
8 0

nope not really sorry

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If 2 objects have the same momentum which statement is true
Vika [28.1K]

C is the correct option

As momentum = (mass) (velocity)

P=mv

So the object having greater mass has less velocity and the object having smaller mass will move faster. But product of mass and velocity is same to each other.

5 0
3 years ago
The buoyant force on an object fully submerged in a liquid depends on (select all that apply)
Natasha_Volkova [10]
2. The object's volume.
 3. The density of the liquid.  
Remember what the buoyant force is. It's the lifting force caused by the displacement of a fluid. I'm using the word fluid because it can be either a liquid or gas. For instance a helium balloon floats due to the buoyant force exceeding the mass of the balloon. So let's look at the options and see what's correct. 
 1. Object's mass
 * This doesn't affect the buoyant force directly. It can have an effect if the object's mass is lower than the buoyant force being exerted. Think of a boat as an example. The boat is floating on the top of the water. If cargo is loaded into the boat, the boat sinks further into the water until the increased buoyant force matches the increased mass of the boat. But if the density of the object exceeds the density of the fluid, then increasing the mass of the object will not affect the buoyant force. So this is a bad choice. 
 2. The object's volume.
 * Yes, this directly affects the buoyant force. So this is a good choice. 
 3. The density of the liquid.
 * Yes, this directly affects the buoyant force. You can drop a piece of iron into water and it will sink. You could also drop that same piece of iron into mercury and it will float. The reason is that mercury has a much higher density than water. So this is a good choice. 
 4. Mass of the liquid
 * No. Do not mistake mass for density. As a mental exercise, imagine the buoyant force on a small piece of metal dropped into a swimming pool. Now imagine the buoyant force on that same piece of metal dropped into a lake. In both cases, the buoyant force is the same, yet the lake has a far greater mass of water than the swimming pool. So this is a bad choice.
8 0
3 years ago
Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of
marusya05 [52]

Answer:

a) \dot m = 6.878\,\frac{kg}{s}, b) T = 104.3^{\textdegree}C, c) \dot S_{gen} = 11.8\,\frac{kW}{K}

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The mass flow rate is:

\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

P = 7000\,kPa

T = 500^{\textdegree}C

h = 3411.4\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

State 2s - Liquid-Vapor Mixture

P = 100\,kPa

h = 2467.32\,\frac{kJ}{kg}

s = 6.8000\,\frac{kJ}{kg\cdot K}

x = 0.908

The isentropic efficiency is given by the following expression:

\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}

The real specific enthalpy at outlet is:

h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})

h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )

h_{2} = 2684.46\,\frac{kJ}{kg}

State 2 - Superheated Vapor

P = 100\,kPa

T = 104.3^{\textdegree}C

h = 2684.46\,\frac{kJ}{kg}

s = 7.3829\,\frac{kJ}{kg\cdot K}

The mass flow rate is:

\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}

\dot m = 6.878\,\frac{kg}{s}

b) The temperature at the turbine exit is:

T = 104.3^{\textdegree}C

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0

\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})

\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )

\dot S_{gen} = 11.8\,\frac{kW}{K}

4 0
4 years ago
What happens to the distance physics help please? The frequency of a wave increases. What happens to the distance between succes
aliya0001 [1]
For any periodic wave 
<span>v = f λ </span>

<span>where </span>
<span>v = velocity </span>
<span>f = frequency </span>
<span>λ = wavelength (distance between 2 successive crests) </span>

<span>This means that </span>
<span>λ = v/f </span>

<span>Assuming that v stays the same while f increases, λ must DECREASE.

I hope this helps

</span>
4 0
3 years ago
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the par
nikdorinn [45]

Answer:

Explanation:

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

7 0
4 years ago
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