Explanation:
The temperature of the solar nebula was decreasing as it moved away from its center. Therefore, only heavy elements could condense in the inner solar system and terrestrial planets could not form with light elements, such as gases. In the outer solar system, the Jovian planets formed mostly with gases, since temperatures were too low to allow rocky compositions.
Answer:
1. v = 30 m/s
2. v = 5 m/s
3. f = 40 Hz
4. f = 400 Hz
5. f = 300 Hz
6. λ = 0.772 m
7. λ = 0.386 m
8. λ = 0.625 m
9. v = 100 m/s
10. v = 50 m/s
Explanation:
The relationship between frequency, wavelength, and speed of a wave is given by the following formula:

where,
v = speed of wave
f = frequency of wave
λ = wavelength
1.
f = 100 Hz
λ = 0.3 m
Therefore,
v = (100 Hz)(0.3 m)
<u>v = 30 m/s</u>
<u></u>
2.
f = 50 Hz
λ = 0.1 m
v = (50 Hz)(0.1 m)
<u>v = 5 m/s</u>
<u></u>
3.
v = 20 m/s
λ = 0.5 m

<u>f = 40 Hz</u>
<u></u>
4.
v = 80 m/s
λ = 0.2 m

<u>f = 400 Hz</u>
<u></u>
5.
v = 120 m/s
λ = 0.4 m

<u>f = 300 Hz</u>
<u></u>
6.
v = 340 m/s
f = 440 Hz

<u>λ = 0.772 m</u>
<u></u>
7.
v = 340 m/s
f = 880 Hz

<u>λ = 0.386 m</u>
<u></u>
<u></u>
8.
v = 250 m/s
f = 400 Hz

<u>λ = 0.625 m</u>
<u></u>
9.
f = 50 Hz
λ = 2 m
v = (50 Hz)(2 m)
<u>v = 100 m/s</u>
<u></u>
10.
f = 100 Hz
λ = 0.5 m
v = (100 Hz)(0.5 m)
<u>v = 50 m/s</u>
Given:
Shaft Power, P = 7.46 kW = 7460 W
Speed, N = 1200 rpm
Shearing stress of shaft,
= 30 MPa
Shearing stress of key,
= 240 MPa
width of key, w = 
d is shaft diameter
Solution:
Torque, T = 
where,

= 59.365 N-m
Now,


d = 0.0216 m
Now,
w =
=
= 5.4 mm
Now, for shear stress in key
= 
we know that
T =
= F. 
⇒
= 
⇒
= 
length of the rectangular key, l = 4.078 mm
Electron is a fundamental Particle by JJ Thomson so it was truly difficult to say that electron is a negative charge
Explanation:
P₁ = P₂ + ρgh
g is the acceleration due to gravity
ρ is the density of the fluid
h is the depth of the fluid
P₁ is the pressure at that depth
P₂ is the pressure at the surface
P₁ and P₂ can either be absolute pressures or gauge pressures, but they must match.
For example, if you wanted to find the <em>absolute</em> pressure at the bottom of an <em>open</em> tank, you would use P₂ = Patm = 14.7 psi or 101.3 kPa.
If instead you wanted to find the <em>gauge</em> pressure, you would use P₂ − Patm = 0 psi or 0 kPa.
If the tank is sealed and pressurized, you would use the P₂ of the tank.