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3 years ago

Which of the following does light demonstrate?

Physics

2 answers:

pogonyaev3 years ago

7 0

Answer:

the answer is A

Explanation:

nataly862011 [7]3 years ago

4 0

Answer:

Explanation:

Light has both wave and particle characteristics. It behaves as wave in examples such as Reflection, refraction, diffraction, interference. It behaves as a particle in photoelectric effect, compton's effect etc. Light is an example of wave-particle duality.

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A bird flies from the South Pole to the North Pole. Part of the journey is 1000 miles that takes 2 weeks. What is the bird’s vel

viktelen [127]

1000 miles = 1610km = 1.61x10^6m
2 weeks = 14 days = 14x24x1440

V=d/t = 1.61x10^6/14x24x1440
= 3.33m/s

6 0

3 years ago

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Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V

From here, we solve for mass:

m = d * V (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>

4 0

3 years ago

A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to

NikAS [45]

Answer:

The dependent variable is academic performance

The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

Explanation:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

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3 years ago

A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca

Mariulka [41]

i believe it is C....tell me if im right please<3

4 0

3 years ago

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