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3 years ago

13

4. In not less than 150 words, write your substantial insights on what you have learned in the past modules in relevance to what

is happening now.

1 answer:

Kitty [74]3 years ago

8 0

Answer:

“I think it’s about putting yourself in the students’ shoes and seeing how a first-time student, maybe someone who hasn’t even taken chemistry before, is looking at it.”

—Valerie Taraborelli, undergraduate chemistry student, University of Arizona1

“In some ways, I think the people who are the most successful as teachers are the ones who are able to remember what it was like being uncertain and not knowing. When you become an expert, things are easy. So the idea is to try and see where [students] are coming from and why they’ve developed this misconception and what you can do to specifically address it.”

—Dee Silverthorn, biology professor, University of Texas2

<h2>Hope this helps !! </h2>

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Mark walked 2 miles east than 1 mile north how would you determine his total displacement

Yuri [45]

They could determine it by counting their total amount of miles mark went by the directions he went to walk those miles.

4 0

3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0

4 years ago

Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average o

Setler79 [48]

Answer:

0.07 m

Explanation:

$L_0$ = Initial length = 1 km = 1000 m

$\Delta T$ = Change in temperature = 1.00°C

$\alpha$ = Coefficient of linear thermal expansion

Volumetric coefficient of expansion of water

$\beta=210\times 10^{-6}^{\circ}C\\\Rightarrow \beta=3\alpha\\\Rightarrow \alpha=\dfrac{\beta}{3}\\\Rightarrow \alpha=\dfrac{210\times 10^{-6}}{3}\\\Rightarrow \alpha=70\times 10^{-6}\ ^{\circ}C$

Change in length is given by

$\Delta L=L_0\alpha \Delta T\\\Rightarrow \Delta L=1000\times 1\times 70\times 10^{-6}\\\Rightarrow \Delta L=0.07\ m$

The change in length is 0.07 m

3 0

3 years ago

One of the great upcoming sports in the Olympics is the sport of curling. Write a brief essay on the uses of momentum collisions

Svetllana [295]

The Olympic sport of curling is one that is practically designed to show Physics in motion. Curling is a sport in which two teams alternate sliding smoothed stone pucks down an ice rink court with the intent to seat their stone closest to the center of the target (called the house). Each team has eight stones, meaning that the team that goes second has the (could be) massive advantage of sending the last stone.

The mass of the stone is important in that the more massive a stone (m) and the speed at which it travels (v) dictates it's momentum (momentum=mxv). As the curling stone slides down the ice (which is relatively frictionless unless acted upon by other players or objects) and having inertia, continues in it's straight course (again, unless acted upon by outside forces). If the stone hits another stone, it transfers some of its momentum in an elastic collision to that stone and the original stone is deflected in a calculable manner.

Collisions are used in the game to either clear opponent's stones from the house or out of their defensive positions, or to make adjustments to one's stones present in the house, all based on the momentum of the moving stone, and its transference.

6 0

3 years ago

Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the

Leona [35]

Answer:

15 deg

Explanation:

Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .

Given:

For first ball, θ1 = 75◦

initial velocity for both the balls, u = 27.2 m/s

for second ball, θ2 = ?

since distance covered by both the balls is same.

Therefore,..

R1=(u^{2} sin2\alpha _{1}) /g[/tex]

the range for the first ball

the range for the second ball

R2=(u^{2} sin2\alpha _{2}) /g[/tex]

(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]

sin2\alpha _{2})=sin2\alpha _{1})

$2\alpha _{2}$ =sin^-1(sin2\alpha _{1})

$\alpha _{2}$ =1/2sin^-1(sin2\alpha _{1})

$\alpha _{2}$ =

15 deg

8 0

4 years ago