If Sienna is using the gymnasium floor to cool off, which position will lower her skin temperature the quickest?

Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?

galben [10]

Answer:

See explanation

Explanation:

First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.

"The meniscus is the curve seen at the top of a liquid in response to its container. The meniscus can be either concave or convex, depending on the surface tension of the liquid and its adhesion to the wall of the container".

Now, according to this definition, and for water, the reading of the volume must be donde at the bottom of the curve of the meniscus. This is because the water gives a concave curve.

If you read it and matches the height of water, you are getting two results:

One, get an accurate value or volume, because it's been done at eye level.

The second fact is that when you do the reading this way, The total pressure is made equal to the atmospheric pressure by adjusting the height of the cylinder until the water level is equal.

8 0

2 years ago

Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lisabon 2012 [21]

As per kinematics equation we are given that

$v^2 = v_o^2 + 2ax$

now we are given that

a = 2.55 m/s^2

$v_0 = 21.8 m/s$

$v = 0$

now we need to find x

from above equation we have

$0^2 = 21.8^2 + 2(2.55)x$

$0 = 475.24 + 5.1 x$

$x = 93.2 m$

so it will cover a distance of 93.2 m

7 0

2 years ago

Read 2 more answers

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

Initial Condition

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

Boundary Conditions

At x=0

$T(0,t)=T_o$

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

At x=L

$-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

2 years ago

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver

Akimi4 [234]

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = $\frac{distance}{time}$

Solution:

The speed of the ball is given by,

speed = $\frac{distance}{time}$

Thus,

Time = $\frac{distance}{speed}$

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

3 0

2 years ago