How many millimeters are in 8 meters?

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$

where $h_1$ and $h_2$ are the height of the column of oil and the unkown liquid, respectively. Writing for $\gamma_{unk}$ , we have

$\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.$

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0

3 years ago

Define the strip attached to the magnet?

Aleonysh [2.5K]

Answer:

A stripe of magnetic information that is affixed to the back of a plastic credit or debit card.

5 0

3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

4 years ago

a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin

Zielflug [23.3K]

Answer:

a) 600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

$Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m$

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

$Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx$

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

$Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}$

7 0

3 years ago

Which states of matter have particles that move independently of one another with very little attraction

Crazy boy [7]

The state of matter that the particles move independently of one another with very little attraction is, I believe, gas

6 0

3 years ago