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3 years ago

11

The wavelengths of light emitted by an excited atom are determined by the:

2 answers:

Natalka [10]3 years ago

7 0

Difference in energy between the exicted states and ground states

Murrr4er [49]3 years ago

3 0

The best and most correct answer among the choices provided by your question is the first choice or letter A.

<span>The wavelengths of light emitted by an excited atom are determined by the: differences in energy between the excited states and ground state.
</span>

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While watching a recent science fiction movie, one Klingon spaceship blows up a Droid spaceship with a laser gun. The Klingon cr

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D. is the right answer because his pressure is very bad out there in the air. good luck

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3 years ago

Read 2 more answers

Explain how the fixed points are used when calibrating a thermometer.

8090 [49]

<u>Answer:</u>

First, the thermometer is dipped into boiling water, and the mercury inside the thermometer rises to a high level, called the boiling point. This level is then marked as 100°C. The thermometer is then dipped into melting ice, which causes the mercury level to fall to a point called the ice point. This point is then marked as 0°C. The length of the thermometer from the 0°C mark to the 100°C point is then divided into 100 equal sections, and the rest of the levels are marked accordingly.

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2 years ago

What is the resultant force of 500g on abject accelerating at 5m/s2

max2010maxim [7]

Answer: 2.5 N

Explanation:

m = 500g = 0.5Kg

a = 5m/s2

F = ma = 0.5 x 5 = 2.5 N

6 0

3 years ago

A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi

sleet_krkn [62]

Answer:

Equilibrium temperature will be $T=52.2684^{\circ}C$

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So $2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)$

$0.334T-3.67=1688-32.031T$

$T=52.2684^{\circ}C$

5 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

b)

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

3 years ago