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How did the temperature structure of the solar nebula determine planetary composition?

STALIN [3.7K]

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.

3 0

3 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

aivan3 [116]

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:

$v = f\lambda$

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)

v = 30 m/s

2.

f = 50 Hz

λ = 0.1 m

v = (50 Hz)(0.1 m)

v = 5 m/s

3.

v = 20 m/s

λ = 0.5 m

$f = \frac{v}{\lambda} = \frac{20\ m/s}{0.5\ m}$

f = 40 Hz

4.

v = 80 m/s

λ = 0.2 m

$f = \frac{v}{\lambda}=\frac{80\ m/s}{0.2\ m}$

f = 400 Hz

5.

v = 120 m/s

λ = 0.4 m

$f = \frac{v}{\lambda}=\frac{120\ m/s}{0.4\ m}$

f = 300 Hz

6.

v = 340 m/s

f = 440 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{440\ Hz}\\$

λ = 0.772 m

7.

v = 340 m/s

f = 880 Hz

$\lambda = \frac{v}{f}=\frac{340\ m/s}{880\ Hz}\\$

λ = 0.386 m

8.

v = 250 m/s

f = 400 Hz

$\lambda = \frac{v}{f}=\frac{250\ m/s}{400\ Hz}\\$

λ = 0.625 m

9.

f = 50 Hz

λ = 2 m

v = (50 Hz)(2 m)

v = 100 m/s

10.

f = 100 Hz

λ = 0.5 m

v = (100 Hz)(0.5 m)

v = 50 m/s

6 0

3 years ago

A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing

Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, $\tau _{shaft}$ = 30 MPa

Shearing stress of key, $\tau _{key}$ = 240 MPa

width of key, w = $\frac{d}{4}$

d is shaft diameter

Solution:

Torque, T = $\frac{P}{\omega }$

where,

$\omega = \frac{2\pi N}{60}$

$T = \frac{7460}{\frac{2\pi (1200 )}{60}}$ = 59.365 N-m

Now,

$\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}$

$30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}$

d = 0.0216 m

Now,

w = $\frac{d}{4}$ = $\frac{0.02116}{4}$ = 5.4 mm

Now, for shear stress in key

$\tau _{key}$ = $\frac{F}{wl}$

we know that

T = $F \times r$ = F. $\frac{d}{2}$

⇒ $\tau _{key}$ = $\frac{\frac{T}{\frac{d}{2}}}{wl}$

⇒ $240\times 10^{6}$ = $\frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}$

length of the rectangular key, l = 4.078 mm

7 0

3 years ago

Read 2 more answers

Is it 100% sur than electron have negative charg and proton have pasitive????

aev [14]

Electron is a fundamental Particle by JJ Thomson so it was truly difficult to say that electron is a negative charge

4 0

4 years ago

In regards to Pressure ( Ch. Static Fluids - Introductory Physics)

ehidna [41]

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 absolute pressure at the bottom of an open tank, you would use P₂ = Patm = 14.7 psi or 101.3 kPa.

If instead you wanted to find the gauge 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.

3 0

3 years ago