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kogti [31]
3 years ago
8

What is a simple machine that has two or more simple machines working together

Physics
1 answer:
zmey [24]3 years ago
3 0

Compound machines have two or more simple machines

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Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t
Vsevolod [243]

Answer:

8.2

Explanation:

8 0
3 years ago
What is the measurement of six centimeters equal to?
AlladinOne [14]
Six centimeters equal to about two inches
8 0
3 years ago
What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?
laila [671]

Answer:

Work done, W = 750 joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

It moves to a distance of, d = 15 meters

We need to find the work done on an object. We know that the product of force and distance covered is called the work done. As the force and the displacement are in same direction. So,

W=F\times d

W=50\ N\times 15\ m

W = 750 joules

So, the work done on an object is 750 joules. Hence, this is the required solution.

4 0
3 years ago
What are some examples for number 4?
lidiya [134]

I would love to help you, but the picture is in a difficult position to read.

5 0
3 years ago
A 4-L pressurecooker has an operating pressure of175 kPa. Initially, one-half of the volume is filled with liquid and the other
Soloha48 [4]

To solve this problem, it is necessary to apply the concepts related to the Energy balance and the mass balance that allow us to find in each state the data necessary to find the total Power of the system through heat exchange.

From the tables of properties of the Water it is possible to obtain at the given pressures the values of the specific volume, the specific energy and the specific enthalpy.

Given these pressures then we have to

P_1 = 174kPA

\Rightarrow v_f = 0.001057m^3/kg\\\Rightarrow v_g = 1.0037m^3/kg\\\Rightarrow u_f = 486.82kJ/kg\\\Rightarrow u_g = 2524.5kJ/kg

P_2 = 175kPa \rightarrow Saturated vapor

\Rightarrow v_2 = v_g = 1.0036 m^3/kg\\\Rightarrow v_2 = u_g = 2524.5kJ/kg

P_e = 175kPa \rightarrow Saturated vapor

V = 4L

Considering the process performed, the kinetic and potential energy can be neglected as well as the work involved by specific interactions in the system. Although it is an unstable process it can be treated as a uniform flow process. Considering these expressions we can perform a mass balance for which

m_{in}-m_{out} = \Delta m_{system} \rightarrow m_e = m_1-m_2

Similarly through the energy balance you can get that

E_{in}-E_{out} = \Delta E_{system}

Q_{in} -m_eh_e = m_2u_2-m_1u_1 \Rightarrow Since W=ke=pe=0

The initial mass, initial internal energy, and final mass in the tank are

m_1 = mf_+m_g = \frac{V_f}{v_f}+\frac{V_g}{v_g}

m_1 = \frac{0.002m^3}{0.001057m^3/kg}+\frac{0.002m^3}{1.0036m^3/kg}

m_1 = 1.893+0.002= 1.895Kg

At the same time the internal energy can be defined from the mass in state 1 as,

U_1 = m_1 u_1

U_1 = m_fu_f+m_gu_g

U_1 = 1.893*486.82+0.002*2524.5

U_1 = 926.6kJ

The calculation of mass in state 2 can be defined as

m_2 = \frac{V}{v_2} = \frac{0.004m^3}{1.0037m^3/kg}

m_2 = 0.004Kg

Then from the mass and energy balances,

m_e = m_1-m_2\\m_e = 1.895-0.004\\m_e = 1.891Kg

In this way the calculation of the heat of entry would be subject to

Q_{in} = m_eh_e+m_2u_2-m_1u_1

Q_{in} = 1.891*2700.2+0.004*2524.5-926.6

Q_{in} = 4188kJ

Therefore the Power would be given as

\dot{Q} = \frac{Q}{\Delta t} = \frac{4188kJ}{3600s} = 1.163kW

Therefore the highest rate of heat transfer allowed is 1.163kW

3 0
3 years ago
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