Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
<em>There are some placeholders in the expression, but they can be safely assumed</em>
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
<u>Sinusoidal Waves
</u>
An oscillating wave can be expressed as a sinusoidal function as follows

Where



The voltage of the question is the sinusoid expression

(a) By comparing with the general formula we have


(b) The period is the reciprocal of the frequency:


Converting to milliseconds

(c) The amplitude is

(d) Phase angle:

It confirmed medeleeve's hypothesis (prediction) and showed the use of his table
Answer:
Explanation:
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>FALSE. </em>The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.
The energies of atoms are not quantized. <em>FALSE. </em>The energies of the atoms are in specific levels.
When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. <em>FALSE. </em>During absorption, a specific wavelength of light is absorbed, not emmited.
Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. <em>TRUE. </em>Again, you can observe just the transition due the change of energy of an electron in the quantized energy level
When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. <em>TRUE. </em>The electron decreases its energy releasing a specific wavelength of light.
The energies of atoms are quantized. <em>TRUE. </em>In fact, the energy of all subatomic, atomic, and molecular particles is quantized.
Answer:
740 N
Explanation:
We are given that
Radius,r=0.3 m
Torque,
We have to find the magnitude of the static frictional force.
According to question
Torque by engine=Torque by static friction




Hence, the magnitude of static frictional force=740 N