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3 years ago

An atom's emission of light with a specific amount of energy confirms that

Physics

1 answer:

kompoz [17]3 years ago

5 0

Answer:

An atom’s emission of light with a specific amount of energy confirms that electrons emit and absorb energy based on their position around the nucleus. The light emitted from an electron is a result of the electron's quantum jumps/leaps ( atomic electron transitions ) to and from different energy levels.

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What is the gravitational force between mars and Phobos

alina1380 [7]

Answer:

$F=5.16\times 10^{15}\ N$

Explanation:

We have,

Mass of Mars is, $m_M=6.42\times 10^{23}\ kg$

Mass of its moon Phobos, $m_P=1.06\times 10^{16}\ kg$

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

$F=G\dfrac{m_Mm_P}{d^2}$

Plugging all values, we get :

$F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N$

So, the gravitational force is $5.16\times 10^{15}\ N$ .

8 0

4 years ago

HELP PLZZZ!!!! Hurry

Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3. $\bar 6$

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7. $\bar 3$

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet, the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

8 0

3 years ago

Read 2 more answers

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

$f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f$

Which for our values means (we do not need the value of n, that is, which harmonics are the frequencies given):

$f=f_{n+1}-f_n=480Hz-400Hz=80Hz$

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

$f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}$

So the tension is:

$T=\mu(2Lf)^2$

Which for our values is:

$T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N$

6 0

3 years ago

When the position of the mass is farthest from the equilibrium position, what is the velocity of the mass?

Llana [10]

At the most distant point, the size of the speed is zero (0 m/s). This is a direct result of preservation of vitality. PE = KE. The most distant far from the harmony position is the maximum PE. Hence it can have no KE. No KE implies no speed since KE = .5mv2

5 0

3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago