Energy flow in an ecosystem is best represented by

When two bodies are charged, the total charge before and after charging remains the same because of:

Oduvanchick [21]

Answer:

b. Conservation of charges

Explanation:

5 0

4 years ago

An undamped 2.47 kg2.47 kg horizontal spring oscillator has a spring constant of 32.8 N/m.32.8 N/m. While oscillating, it is fou

olganol [36]

Answer:

0.631 m

6.53315 J

Explanation:

m = Mass = 2.47 kg

v = Velocity = 2.30 m/s

k = Spring constant = 32.8 N/m

A = Amplitude

In this system the energy is conserved

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{2.47\times 2.3^2}{32.8}}\\\Rightarrow A=0.631\ m$

The amplitude is 0.631 m

Mechanical energy is given by

$E=\dfrac{1}{2}mv^2\\\Rightarrow E=\dfrac{1}{2}2.47\times 2.3^2\\\Rightarrow E=6.53315\ J$

The mechanical energy is 6.53315 J

5 0

3 years ago

A proton is released in a uniform electric field, and it experiences an electric force of 2.36×10−14 N toward the south. (a)What

qwelly [4]

Answer:

a) E = 1.47 × 10^5 N/C

b) south

Explanation:

The magnitude of an electric field can be defined mathematically as;

E = F/q ........1

Where,

E = magnitude of the electric field

F = electric force

q = charge on the proton

Given;

F = 2.36 × 10^-14 N

Note that charge on a proton is known as Qp = 1.602 × 10^-19 C

q = 1.602 × 10^-19 C

Substituting into equation 1, we have;

E = 2.36 × 10^-14 N/1.602 × 10^-19 C

E = 1.47 × 10^5 N/C

b) The direction of the electric field;

From equation 1

E = F/q ........1

since both electric field and electric force are vector quantity and q is a positive charge (constant), then both the electric field and electric force would be parallel to each other. Therefore the electric field is directed to the south also.

(When a vector is multiplied by a positive constant the direction remains the same)

7 0

4 years ago

An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's

ivolga24 [154]

1) Focal length

We can find the focal length of the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$ (1)
where
f is the focal length
$d_o$ is the distance of the object from the mirror
$d_i$ is the distance of the image from the mirror

In this case, $d_o = 18 cm$ , while $d_i=-4 cm$ (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
$\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}$
from which we find
$f=- \frac{36}{7} cm=-5.1 cm$

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
$r=2f=2 \cdot 5.1 cm=10.2 cm$

3 0

4 years ago

Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f

navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

ΔE = $E_{nf}$ - E₀ = - k²e² / 2m (1 / $n_{f}$ ²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

E = h f = h c / λ

h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

1 / λ = Ry (1/ $n_{f}$ ² - 1 / n₀²)

Let's apply these equations to our case

λ = 821 nm = 821 10⁻⁹ m

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

$E_{nf}$ - E₀ = -13.606 (1 / $n_{f}$ ² - 1 / n₀²) [eV]

Let's look for the energy of some levels

n $E_{n}$ (eV) $E_{nf}$ - E $E_{ni}$ (eV)

1 -13,606 E₂-E₁ = 10.20

2 -3.4015 E₃-E₂ = 1.89

3 -1.512 E₄- E₃ = 0.662

4 -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0

3 years ago