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3 years ago

1) Manipulate is to measure as

1 answer:

3 0

Answer: The variable that you manipulate is called the independent variable. The variable that you measure is called the dependent variable.

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Remember to identify all your data, write the equation, and show your work.

Paraphin [41]

Answer:

I believe it's 8.09 seconds, but I'm rusty on my physics.

Explanation:

The equation for solving the time it takes for an object to fall is $\sqrt{2d/g}$

So multiply the distance times 2, and you get 642 meters. Then you divide by gravities acceleration constant, 9.8, and you get 65.51. Finally, $\sqrt{65.51}$ , and you get 8.09 seconds.

I pulled the equation off of wikipedia and I'm unsure if it's the correct one, so hopefully this is correct. :/

6 0

3 years ago

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Anestetic [448]

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

8 0

3 years ago

Read 2 more answers

Which of the following body systems work together to get sells the oxygen they need to carry at the cellular respiration

djyliett [7]

The cells in your body help get oxygen to cellular respiration.

8 0

3 years ago

PLEASE HELP IN ONE MINUTE

Greeley [361]

I think it’s 8 hours. I’m sorry if I’m wrong.
I just did 400 divided by 50

4 0

3 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

3 years ago