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tangare [24]
3 years ago
6

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?

Physics
2 answers:
matrenka [14]3 years ago
6 0
"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>
Bas_tet [7]3 years ago
5 0

Answer:

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know that the object position is at center of curvature of the mirror

So here the distance is given as

d = R = 2f

now by mirror formula of relation between distance of image and distance of object

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{2f} = \frac{1}{f}

d_i = 2f

so image will form at the position of object itself

Now we will say that the magnification of the object will be

M = -\frac{d_i}{d_o} = -1

so image will be upside down and same as that the height of object

so correct answer would be

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

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DNA’s diameter is 0.000000002 meters. What is this measurement in scientific notation? a. 2.0 x 109 m
NARA [144]

Answer:

0.000000002 m=2.0*10⁻⁹ m

Explanation:

Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.

A number written in scientific notation has the form:

a*10ⁿ

where:

  • the coefficient a has a value such that 1 ≤ a <10
  • n is an integer.  Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.

So to write the number 0.000000002 in scientific notation, the following steps are performed:

  • The decimal point is moved to the right as many spaces until it reaches the right of the first digit.
  • This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
  • The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.

So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>

3 0
3 years ago
A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?
Vitek1552 [10]

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

8 0
3 years ago
B. On a separate sheet of paper, describe the different ways of generating electric power. ​
Afina-wow [57]

Answer:

These all different sources of energy add to the store of electrical power that is then sent out to different locations via high powered lines. It is the energy from the sun that is harnessed using a range of technologies such as solar heating, solar architecture, photovoltaics, and artificial photosynthesis.

Hope it helps PLS MARK ME AS BRAINLIST I BEG YOU thanks :)

4 0
2 years ago
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
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