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3 years ago

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In an experiment, the explanation of the expected outcome, based on research, is the _______________.

1 answer:

vladimir1956 [14]3 years ago

7 0

C. Hypothesis
The hypothesis is presented as an explanation of the observed results.

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A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

mario62 [17]

Answer:

$0.2$ mm

Explanation:

As we know that

$Y = \frac{m\lambda * D}{d}$

where

m represents the order of minimum

y represents the distance on the screen of the minimum from central axis

λ is the wavelength of the light

D is the distance between screen-to-slit

d represents the width of the slit

For first minima

$y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}$

For second minima

$y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}$

$Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm$

3 0

3 years ago

calculate the energy spent on spraying a drop of mercury of 1 cm radius into 10^6 droplets all of same radius. surface tension o

WARRIOR [948]

Answer:

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

Explanation:

<em>Given:</em>

<em>Radius of mercury = 1cm initially ;</em>
<em>split into $10^{6}$ drops ;</em>

Thus, volume is conserved.

i.e ,

$\frac{4}{3} \pi R_{o}^{3} = 10^{6}*\frac{4}{3} \pi R_{n}^{3}\\R_{n}=\frac{R_{o}}{10^{2}} = \frac{1cm}{100} = 0.01 cm$

Energy of a droplet = $T$ Δ $A$

Where ,

<em>T is the surface tension </em>
<em>ΔA is the change in area</em>

Initial energy $E_{i} = T*A_{i}\\= 0.0035 * 4 *\pi *0.01^{2}\\=4.398*10^{-6}J$

Final energy $E_{f}=10^{6}*T*A_{f}\\=10^{6}*0.0035*4*\pi *(0.0001)^2\\=4.39823*10^{-4}$

∴ .Energy spent on spraying = $=E_{f}-E{i}\\=(439.823-4.39823)*10^{-6}\\=4.3542*10^{-4}J$

ANS : .Energy spent on spraying = $4.3542*10^{-4}J$

6 0

3 years ago

Once the roller coaster train gets closer to the bottom of the hill, its kinetic energy increases to 1,100 J, and its potential

Paladinen [302]

Answer:

2000 For The First/900 For The Second/And 2000 For The Third/JUST TOOK QUIZ

Explanation:

5 0

3 years ago

Read 2 more answers

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

3 years ago

Alex drives 1900 meters in 200 s. Assuming he does not speed up or slow down, what is his speed?

pochemuha

Alex is driving at 21.2508882 miles per hour

8 0

1 year ago