Answer:
T=280.41 °C
Explanation:
Given that
At T= 24°C Resistance =Ro
Lets take at temperature T resistance is 2Ro
We know that resistance R given as
R= Ro(1+αΔT)
R-Ro=Ro αΔT
For copper wire
α(coefficient of Resistance) = 3.9 x 10⁻³ /°C
Given that at temperature T
R= 2Ro
Now by putting the values
R-Ro=Ro αΔT
2Ro-Ro=Ro αΔT
1 = αΔT
1 = 3.9 x 10⁻³ x ΔT
ΔT = 256.41 °C
T- 24 = 256.41 °C
T=280.41 °C
So the final temperature is 280.41 °C.
Answer:
λ = 5.656 x 10⁻⁷ m = 565.6 nm
Explanation:
Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:
where,
λ = wavelength = ?
Δx = fringe spacing = 1.6 cm = 0.016 m
L = Distance between slits and screen = 4.95 m
d = slit separation = 0.175 mm = 0.000175 m
Therefore,
<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>
Answer:
Chemical composition, Temperature, Radial velocity, Size or diameter of the star, Rotation.
Explanation:
Elemental abundances are determined by analyzing the relative strengths of the absorption lines in the spectrum of a star.
The Spectral class to which the star belongs gives the information related to the temperature of the star. It is the spectral lines that determine the spectral class O B A F G K M are the spectral classes.
By measuring the wavelengths of the lines in the star's spectrum gives the radial velocity. Doppler shift is the method used to find the radial velocity.
A star can be classified as a giant or a dwarf . A giant star will have narrow width spectral lines whereas a dwarf star has wider spectral lines.
Broadening of the spectral lines will determine the star's rotation.
Answer:
Twice
Step-by-Step Explanation:
Time between 7:00 PM and 1:00 AM: 6 hours
Distance: 4818km
Since the distance is 4818km, and the time is 6 hours, you divide 4818 by 6.
803.0000015999 km/h.
The average speed is 803 km/h
Which considering the ideal case scenario if the plane starts at 0 reaches the speed of 803 and the end reduces its speed from 803 to 0. This means we have come across the value of 800 at least twice. Hence, the plane was travelling at a speed of 800 km/h at least 2 times.
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
