root mean square<span>= square root of ( 3RT/M)
R = 8.314 J/K/mole
T = 25 + 273 = 298 K
M = molecular mas of N2 in kg = 28 X 10^-3 kg
put values...
</span><span> root mean square</span> = square root of ( 3 X 8.314 X 298/28 X 10^-3)
= square root of ( 265454.143)
= 515.2 m/s
so option A is right
hope this helps
Answer:
2917.4 m/s
Explanation:
From the question given above, the following data were:
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
Radius (r) of the Moon = 1737 Km
Escape velocity (v) =?
Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:
Gravitational acceleration of the earth = 9.8 m/s²
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
= 0.25 × 9.8 = 2.45 m/s²
Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:
1 Km = 1000 m
Therefore,
1737 Km = 1737 Km × 1000 m / 1 Km
1737 Km = 1737000 m
Thus, 1737 Km is equivalent to 1737000 m
Finally, we shall determine the escape velocity of the rocket as shown below:
Gravitational acceleration of the Moon (g) = 2.45 m/s²
Radius (r) of the moon = 1737000 m
Escape velocity (v) =?
v = √2gr
v = √(2 × 2.45 × 1737000)
v = √8511300
v = 2917.4 m/s
Thus, the escape velocity is 2917.4 m/s
(a) -48.0 cm, diverging
We can use the lens equation:
where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:
The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:
where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':
and the image is erect, since the sign is positive.
(c)
See attached picture.
