A hydrogen atom is in its ground state when its orbital electron:

Sugar dissolved in water is an example of?

lbvjy [14]

Answer:

D. Solution

Explanation:

Sugar dissolved in water is an example of solution.

A solution is a homogenous mixture of solutes and solvents.

In a solution the solute particles ae distributed uniformly in the solvents. The solute is the substance and it is the sugar here that is dissolved to make a solution.

The solvent is the water in this instance that helps to dissolve the solute.

5 0

3 years ago

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A Cathode of initial mass 10.00g weigh to 10.05g

Leto [7]

Answer:

i'm sorry i'm not a physics student

6 0

3 years ago

Examine the spectra of the four unknown substances shown below. What can you conclude?

oksian1 [2.3K]

Line spectra are obtained when individual elements are heated using a high-voltage electrical discharge. This heating causes excitation of the element and a subsequent emission of distinct lines of colored light are obtained. Each element has its own unique emission line spectrum; therefore, if any of the tested substances were the same, their spectra would match. However, this is not the case so none of the substances are the same.
hope it helps!

7 0

3 years ago

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

PLEASE HELP!! DUE SOON!! THANKS BRAINLIEST OFFERED!!

pav-90 [236]

Answer:

real, inverted, and smaller than the object

Explanation:

When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.

The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.

7 0

3 years ago