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4 years ago

8

The____of matter depends upon how close the individual particles are together

1 answer:

Ksenya-84 [330]4 years ago

4 0

The (Close) of matter depends upon how close the individual particles are together

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What happens to the width of the central diffraction pattern (in the single slit experiment) as the slit width is changed and wh

Leno4ka [110]

Answer:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

D is distance of screen from source , d is slit width and λ is wavelength of light source.
Here we see , on d getting decreased , width will increase because d is in denominator .

Due to increased width , position of a fringe moves away from the centre.

7 0

3 years ago

You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongs

Fudgin [204]

Answer:

3.6m

Explanation:

if you are at a building that is 46m above the ground, and the professor is 1.80m, the egg must fall:

46m - 1.80m = 44.2m

the egg must fall for 44.2m to land on the head of the professor.

Now, how many time this takes?

we have to use the following free fall equation:

$h=v_{0}t+\frac{1}{2}gt^2$

where $h$ is the height, $v_{0}$ is the initial velocity, in this case $v_{0}=0$ . $g$ is the acceleration of gravity: $g=9.81m/s$ and $t$ is time, thus:

$h=\frac{1}{2}gt^2$

clearing for time:

$2h=gt^2\\\frac{2h}{g}=t^2\\\sqrt{\frac{2h}{g}} =t$

we know that the egg has to fall for 44.2m, so $h=44.2$ , and $g=9.81m/s$ , so we the time is:

$t=\sqrt{\frac{2(44.2m)}{9.8m/s^2} }=\sqrt{\frac{88.4m}{9.81m/s^2} } =\sqrt{9.011s^2}= 3.002s$

Finally, if the professor has a speed of $v=1.2m/s$ , it has to be at a distance:

$d=vt$

and t=3.002s:

$d=(1.2m/s)(3.002s)=3.6m$

so the answer is the professor has to be 3.6m far from the building when you release the egg

7 0

3 years ago

Which change of state takes place when a gas loses energy?

tatuchka [14]

Answer:

Condensation (((((((((((((

7 0

3 years ago

Read 2 more answers

A 70.0 kg base runner moving at a speed of 4.0 m/s begins his slide into second base. The coefficient of friction between his cl

Andre45 [30]

Answer:

B. 560 J

J. 1.2 m

Explanation:

v = Final velocity = 0

u = Initial velocity = 4 m/s

$\mu$ = Coefficient of friction = 0.7

m = Mass of runner = 70 kg

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Kinetic energy is given by

$K=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow K=\dfrac{1}{2}\times 70\times (0^2-4^2)\\\Rightarrow K=-560\ \text{J}$

The mechanical energy lost is 560 J

Acceleration is given by

$a=-\mu g\\\Rightarrow a=-0.7\times 9.81\\\Rightarrow a=-6.867\ \text{m/s}^2$

From kinematic equations we get

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-4^2}{2\times -6.867}\\\Rightarrow s=1.165\approx 1.2\ \text{m}$

The runner slides for 1.2 m

4 0

3 years ago

Ron weighs a bottle with a mass of 0.5 kilograms on a spring scale. The spring stretches by 1 centimeter. He then weighs a secon

Yuri [45]

Answer:

1 kg

Explanation:

Applying,

Hook's law,

F = ke................. Equation 1

Where F = force, k = force constant of the spring, e = extension.

But from the question,

The weight of the bottle is the force acting on the spring scale

therefore,

mg = ke............ Equation 2

Where m = mass of the bottle, g = acceleration due to gravity.

make k the subject of the equation

k = mg/e............ Equation 3

Given: m = 0.5 kg, e = 1 cm = 0.01 m

Constant: g = 9.8 m/s²

k = (0.5×9.8)/0.01

k = 490 N/m

If the mass of the second bottle is weighed,

given: e = 2 cm = 0.02 m

subtitute into equation 1

m×9.8 = 490×0.02

9.8m = 9.8

m = 9.8/9.8

m = 1 kg.

Hence the mass of the second bottle is 1 kg

5 0

3 years ago