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tangare [24]
3 years ago
11

During one year, eight moose in a population died and two moose were born. Three moose immigrated from another population and fi

ve emigrated to find mates.
What was the population growth during this year?
Physics
2 answers:
galben [10]3 years ago
8 0

-8 hope it helps i just took the test

Gnesinka [82]3 years ago
5 0
The population of moose would be for this year is -8
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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The io
IRISSAK [1]

Answer:

\lambda=25.6nm

Explanation:

The Rydberg formula can be extended for use with any hydrogen-like chemical elements, that is to say with only one electron being affected by effective nuclear charge. So, in this case, we can calculate the wavelenghts of the emitted photons using this formula:

\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where R is the Rydberg constant of the element, Z its atomic number, n_1 is the lower energy level and n_2 the upper energy level of the  electron transition. Recall that the ground state is denoted as n=1.

\frac{1}{\lambda}=1.1*10^7m^{-1}(2)2^2(\frac{1}{1^2}-\frac{1}{3^2})\\\frac{1}{\lambda}=3.91*10^7m^{-1}\\\lambda=2.56*10^{-8}m=25.6nm

7 0
3 years ago
A proton travels with a speed of 1.8×106 m/s at an angle 53◦ with a magnetic field of 0.49 T pointed in the +y direction. The ma
Likurg_2 [28]

Answer:

Magnetic force, F=1.12\times 10^{-13}\ N

Explanation:

It is given that,

Velocity of proton, v=1.8\times 10^6\ m/s

Angle between velocity and the magnetic field, θ = 53°

Magnetic field, B = 0.49 T

The mass of proton, m=1.672\times 10^{-27}\ kg

The charge on proton, q=1.6\times 10^{-19}\ C

The magnitude of magnetic force is given by :

F=qvB\ sin\theta

F=1.6\times 10^{-19}\ C\times 1.8\times 10^6\ m/s\times 0.49\ Tsin(53)

F=1.12\times 10^{-13}\ N

So, the magnitude of the magnetic force on the proton is 1.12\times 10^{-13}\ N. Hence, this is the required solution.

3 0
4 years ago
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

C = \frac{Q}{V}  = e_0\frac{A}{d}

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0
3 years ago
How many electrons can the first shell hold
Naily [24]
First shell hold up to 2 electrons.
8 0
4 years ago
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