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3 years ago

When an electron moves from a lower to a higher energy level

1 answer:

8 0

The electrons in an atom exist in various energy levels. When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released (often as light).

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In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?

gizmo_the_mogwai [7]

It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left

5 0

3 years ago

Read 2 more answers

A high jumper jumps 2.04 m. If the jumper has a mass of 67 kg, what is his gravitational potential energy at the highest point i

Mariulka [41]

Answer: 1339.5 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the jumper as he moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 67kg

g = 9.8m/s^2

h = 2.04 metres

Thus, GPE = 67kg x 9.8m/s^2 x 2.04m

GPE = 1339.5 joules

Thus, the gravitational potential energy at the highest point is 1339.5 joules

3 0

3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

A beam oflight is reflected off a mirror. If the angle of incidence is 30 degrees, according to the Law of Reflection the angle

zaharov [31]

Answer:

30°

Explanation:

According to the second law of reflection, it States that the angle of incidence i is equal to the angle of reflection r.

The angle of incidence is known to be the angle between the incident ray and the normal.

The Angle of reflection is the angle between the reflected ray and the normal.

This normal ray is a ray that is perpendicular to the surface.

According to the question, if the beam of light is reflected off the surface and its angle of incidence is 30°, its angle of reflection will also be 30° i.e i=r = 30°

7 0

3 years ago

3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.

Klio2033 [76]