Answer:
Explanation:
As we stir the pot of soup by a metal spoon which is a conductor, it conducts the heat and starts heating after some time the temperature of the spoon rises and we are not able to hold the spoon and we get burns.
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
Answer:
The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular
Explanation:
The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.
The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.
Answer:
1. 579 x 10 ^-22N
Explanation:
F = kq1q2/r^2
= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2
= 1. 579 x 10 ^-22N
Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
Second minima from center
The distance between the places where the intensity is zero due to the double slit effect
Put the value into the formula
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
