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3 years ago

What causes diffferent air density?

Physics

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

The temperature and gravity both affects the density

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a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago

A bike rider pedals with constant acceleration to reach a velocity of 7.8 m/s over a time of 4.2 s. during the period of acceler

Artyom0805 [142]

To calculate the initial velocity of the bike, we use the following equation

$d=\frac{1}{2} (u+v)t$ .

$u=\frac{2d}{t} -v$

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.

Given, $u =7.8 m/s$ , $d= 19 m$ and $t=4.2 s$ .

Substituting these values in above equation, we get

$u = \frac{2 \times 19}{4.2 \ s} -7.8 m/s = 9.05 \ m/s - 7.8 \ m/s \\\\ u= 1.2 m/s$ .

Thus, the initial velocity of the bike is 1.2 m/s.

3 0

3 years ago

How deep is the outer core beneath the surface

sukhopar [10]

Precisely around 1,800 miles below.

6 0

3 years ago

The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one

marishachu [46]

Answer

given,

I is the loudness of sound

I = 10 Log₁₀ r

r is relative intensity

at when relative intensity is 10⁶

I = 60 dB

how much louder when 100 people would be talking together

I = 10 Log₁₀ r

I = 10 Log₁₀ (10⁶ x 100)

I = 10 Log₁₀ (10⁸)

I = 80 dB

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

5 0

3 years ago

Please help

iVinArrow [24]

Answer is 4,400,000 kg • m/s

4 0

3 years ago