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julia-pushkina [17]

3 years ago

9

What is defined as the isolation and reduction of variables in an experiment in order to understand the impact of a specific var

iable or variables on the outcome of the study?

1 answer:

ehidna [41]3 years ago

7 0

aisnhsjdjxbxbxhxbxjx

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true or false Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer

salantis [7]

Hello Micu212006

Question: <span> Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
</span><span>
Answer: True

Hope This Helps!
-Chris </span>

8 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

What is the moon phase today

Elis [28]

Waxxing Gibbous Phase

4 0

2 years ago

Read 2 more answers

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before

RideAnS [48]

The velocity of pluck 1 is 12 m/s west.

<h3>What is the conservation of momentum?</h3>

The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.

Now given that;

m1u1 + m2u2 = m1v1 + m2v2

(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)

1.5 - 1.2 = 0.1v + 1.5

0.3 - 1.5 = 0.1v

v = -1.2/0.1

v = - 12 m/s

Hence, the velocity of pluck 1 is 12 m/s west.

Learn more about linear momentum:brainly.com/question/27988315

#SPJ1

7 0

1 year ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago