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nasty-shy [4]
3 years ago
5

Please I need help with

Physics
1 answer:
Svetlanka [38]3 years ago
4 0
Of chromosomes each cell type has different genes Althea answer is the three answer in the picture
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How do you compare the energies of the different parts of the electromagnetic spectrum?
faltersainse [42]

The different types of radiation in electromagnetic spectrum are compared by the amount of energy found in the photons.

Radio waves have photons with low energies, microwave photons have a little more energy than radio-waves. Infrared photons still have more energy, then comes visible, ultraviolet, x-rays and the most energetic of all, gamma rays.

The energy associated with electromagnetic radiation is proportional to frequency and inversely proportional to wavelength. So, electromagnetic waves with shorter wavelengths have more energy.

On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light. On the other end of the spectrum are gamma rays with wavelengths billions of times smaller than those of visible light.

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5 0
1 year ago
Properties seen when one substance changes to another are know as ____ properties
maxonik [38]

Answer: chemical property

Explanation:

6 0
3 years ago
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Write about the similarities and differences between kinetic and potential energy. Include specific, real world example in your
lord [1]

Answer:

Energy is transformed from potential to kinetic and vice versa

Explanation:

The energy is transformed from mechanical to kinetic energy when the object changes its position with respect to a reference point, where it loses height but increases its speed. When the object is at maximum height with respect to a reference point, it will have its maximum potential energy value. When the object passes through the reference point it will have potential energy equal to zero, but this energy will become kinetic energy.

The most characteristic and real example is that of a pendulum at one end, as can be seen in the attached image.

When the pendulum is located at the top end, as shown in Figure 1, at that point the maximum potential energy will be held. Then the pendulum is released and when it passes through the reference point and its height is zero, with respect to that point, all potential energy will have become kinetic energy in the same way at this point the maximum speed of the pendulum will be set.

8 0
3 years ago
How do I solve the following problem?
lukranit [14]
Use the impulse-momentum theorem.

Ft = mv

Substitute your known values:

(120kg)(20m/s) = F(1.5)

F= 1600N

Hope this helps!
7 0
3 years ago
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

5 0
3 years ago
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