Ask question

Ask question

All categories

2 years ago

6

A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the

water rises part-way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the water surface on the other side of the U. What is the density of the oil

1 answer:

Doss [256]2 years ago

8 0

Answer:

0.89 g/cm^3 = 890 kg/m^3

Explanation:

Cross sectional area of U-tube ( A ) = 1.00 cm^2

volume of oil ( V ) = 5.00 cm^3

change between top surface = 0.550 cm

height of oil = 5 cm ( volume / area )

height of water = 5 - 0.550 = 4.45 cm

pressure at the oil-water junction = Pressure on the second side of the U-tube at same level

Po * g * Hoil = Pw * g * Hwater

Po * 5 = 1 * 4.45

∴ Density of oil ( Po ) = 4.45 / 5 g/cm^3 = 0.89 g/cm^3

You might be interested in

What is a device that minimizes the force on a microscopic object during a collision?

liraira [26]

An impact which stops a moving object must do enough work to take away its kinetic energy, so extending the distance moved during the collision reduces the impact force.

3 0

2 years ago

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

7 0

2 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

2 years ago

Why do astronauts' weights differ in space?

OLga [1]

Answer:

There is less gravitational force in space.

Explanation:

Gravity doesn't exist as it does on Earth. Earth's gravity is as 6 times stronger as it is on the moon.

7 0

2 years ago

A chemical is always the same thing as an element .<br><br> True <br> False

olya-2409 [2.1K]

That would be false

Hope this helps :)

3 0

2 years ago

Read 2 more answers