Question 8

You come across an open container which is filled with two liquids. Since the two liquids have different density there is a dist

Slav-nsk [51]

Answer:

526.57 Pa

Explanation:

P ( pressure at the bottom of the container) = 1.049 × 10^5 pa

Using the formula of pressure in an open liquid

Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²

Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa

P( atmospheric pressure) = 1.013 × 10^5 Pa

Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g

Subtract each of the pressure from the absolute pressure at the bottom

P(bottom) - atmospheric pressure

(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa

subtract pressure due to water from the remainder

3600 - 2050.29 = 1549.71 Pa

1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g

ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa

4 0

2 years ago

A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.

Bond [772]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

4 0

2 years ago

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

Agata [3.3K]

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

Explanation:

For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

$\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m$

Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

$\boxed{T = 1.5*10^{-6}N}$

or in micro newtons

$T =1.5\mu N$

4 0

3 years ago

Current assets minus current liabilities is known as

levacccp [35]

Answer:

Working capital is a fundamental accounting concept essential to running a business. Essentially, working capital is a company's current assets minus its current liabilities. Current assets are typically those that are highly liquid, such as cash or inventory

Explanation:

i

hope it helped...

3 0

2 years ago

The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

just olya [345]

Answer:

1.99 parsecs.

Explanation:

We have been given that the most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-years.

We know that one light year equals to 0.306601 parsecs. To convert 6.5 light-years to parsecs, we will multiply 0.306601 by 6.5.

$6.5\text{ Light-years}=6.5\times 0.306601\text{ Parsecs}$

$6.5\text{ Light-years}=1.9929065\text{ Parsecs}$

$6.5\text{ Light-years}\approx 1.99\text{ Parsecs}$

Therefore, Luhman 16 is approximately 1.99 parsecs away from the Earth.

5 0

3 years ago