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Alenkasestr [34]

3 years ago

7

What is any factor of an experiment that is not changed called? control, dependent variable ,independent variable

2 answers:

Svetllana [295]3 years ago

8 0

Control is fixed, dependent variable is controlled but changes, the independent is observed

Softa [21]3 years ago

5 0

A control is the factor which is not changed

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an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

$a_c = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

$a_c = \frac{(12\ m/s)^2}{2\ m}$

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

$F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)$

<u>Fc = 504 N</u>

6 0

3 years ago

An embolus can cause which of the following?

QveST [7]

I think it’s C. Stroke if not then D

7 0

3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C

In-s [12.5K]

because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.

4 0

3 years ago

Samantha and Emily are pushing a box of textbooks in the same direction across their classroom. Samantha is applying a force of

velikii [3]

Answer: 20 newtons

Explanation:

Given that:

Force applied by Samantha = 10 newtons

Force applied by Emily = 10 newtons

Direction of both forces = same

Net force of their efforts = ?

Since net force of forces applied in the same direction is obtained by adding up the seperate forces applied, then

Net force = (10 newtons + 10 newtons)

Net force = 20 newtons

Thus, the net force of their efforts is 20 newtons.

3 0

4 years ago