Units and Numbers
In order to determine an object’s speed on a Speed vs. Time graph, the axes MUST be labeled with United and numbers.
Answer:
Explanation:
capacitance of capacitor = ε₀ A / d
ε₀ = 8.85 x 10⁻¹² , A is area of plate and d is plate separation .
= 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 3.4 x 10⁻³
C = 24.73 x 10⁻¹³
capacitance of capacitor after increase in plate separation
= 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 9.5 x 10⁻³
= 8.85 x 10⁻¹³
initial charge = capacitance x potential
= 24.73 x 10⁻¹³ x 7.6 C
potential difference after increased separation
= initial charge / increased capacitance
= 24.73 x 10⁻¹³ x 7.6 / (8.85 x 10⁻¹³)
= 21.23 V .
b ) initial stored energy
= 1/2 C V²
= .5 x 24.73 x 10⁻¹³ x 7.6²
= 714.2 x 10⁻¹³ J
c ) final stored energy
1/2 C V²
= .5 x 8.85 x 10⁻¹³ x 21.23 ²
= 1994.4 x 10⁻¹³ J
d ) work done in separation of plate
=2 x increase in stored energy
= 2 x (1994.4 - 714.2 ) x 10⁻¹³ J
= 2560.4 x 10⁻¹³ J .
The characteristics of the atomic spectrum allow to find the result so that the measured lines help the astronomer is:
- The astronomer can identify the gases that are present in stars using emission lines.
The emission spectrum is the emission of light due to the atomic transition in atoms, consequently the spectrum of each chemical element is unique.
The astronomer was able to identify the chemical elements that are in stars by their emission spectrum.
In the case presented there are two lines in the blue region, a line in the green region and a line in the red region. We know that the most abundant element in stars is hydrogen.
Hydrogen has a series of well-defined spectral lines, the series that is in the visible part of the spectrum is called the Balmer series and is composed of the lines in the table.
Color wavelength (nm)
red 656.3
blue 486.1
violet 434.1
The second element in quantity in stars is helium, which has the spectrum given in the table and in the attached.
color wavelength (nm)
Red 667.8
orange 587.6
green 501.6
Blue 492.2
Blue 447.1
The astronomer compares his measured spectrum with the spectra of these two gases and finds that the element he is measuring is helium.
In conclusion, using the characteristics of the atomic spectra, we can find the result to help the astronomer with the measured lines is:
- The astronomer can identify the gases that are present in stars using emission lines.
Learn more here: brainly.com/question/14819167
Answer:
Explanation:
Given
radius of Planet is equal to radius of Earth
Weight of body on Planet
where m=mass of body
Weight of body on earth
acceleration due to gravity is given by
where G=gravitational constant
M=mass of Planet
r=radius of planet
for earth
for planet
substituting these values in and
divide 1 and 2
Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s
Explanation:
Magnetic force(B) = 4.60×10^-3 T
Electric force(E) = 1.64×10^4 V/m
Both forces having equal magnitude ;
Magnetic force = electric force
qvB = qE
vB = E
v = (1.64×10^4) ÷ (4.60×10^-3)
v = 3.57×10^6 m/s
2.) Assume no electric field
qvB = ma
Where a = v^2 ÷ r
R = radius
a = acceleration
v = velocity
qvB = m(v^2 ÷ R)
R = (m×v) ÷ (|q|×B)
q=1.6×10^-19C
m = 9.11×10^-31kg
R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)
R = 32.5227×10^-25 ÷ 7.36×10^-22
R = 4.42×10^-3m
3.) period(T)
T = (2*pi*R) ÷ v
T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)
T = (27.775×10^-3) ÷(3.57×10^6)
T = 7.78×10^-9 s