Abstract on which beverage has the most gass between coke,orange juice and water

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the

Liono4ka [1.6K]

The energy of the ski lift at the base is kinetic energy:
$K= \frac{1}{2}mv^2$
where m is the mass of the ski lift+the people carried, and $v=15.5 m/s$ is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the height at the top of the hill.

So, since the total energy must conserve, we have
$U=K$
and so
$mgh = \frac{1}{2}mv^2$
from which we find the height:
$h= \frac{v^2}{2g}= \frac{(15.5m/s)^2}{2\cdot 9.81 m/s^2}=12 m$

8 0

3 years ago

Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains

tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v = (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0

4 years ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

3 years ago

Type the correct answer in the box. Express your answer to three significant figures.

Marrrta [24]

The mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

Given;

Mass of calcium nitrate (Ca(NO₃)₂) = 96.1 g

Mass of calcium phosphate=?

The chemical equation is found as;

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO4)₂

3 moles of Ca(NO₃)₂ produces 1 mole of Ca₃(PO4)molar mass of calcium phosphate is 164 g/mol

Moles are found as the ratio of the mass of the compound and its molar mass. Moles of Ca(NO₃)₂ is n.

n = 96.1/164 = 0.5859 moles

Moles of Ca3(PO4)2 produced ;

N=0.0589 ×(1/3) = 0.0196 moles

The molar mass of calcium phosphate is 310 g/mol and the mass of calcium phosphate produced will be;

M=0.0196×310

M = 6.076 g

Hence,the mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

To learn more about the mass, refer to the link;

brainly.com/question/13073862

#SPJ1

5 0

3 years ago

An archery bow is drawn a distance d = 0.29 m and loaded with an arrow of mass m = 0.094 kg. The bow acts as a spring with a spr

dezoksy [38]

Answer:

The arrow will leave the bow with a velocity of 10 m/s.

Explanation:

Hi there!

The potential energy stored in the bow can be calculated using the following equation:

U = 1/2 · k · d²

Where

U = elastic potential energy.

k = spring constant.

d = stretched distance of the bow

Then:

U = 1/2 · 112 N/m · (0.29 m)²

U = 4.7 J

When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:

KE = 1/2 · m · v² = 4.7J

Where:

KE = kinetic energy.

m = mass of the arrow.

v = velocity of the arrow:

Then:

4.7 J = 1/2 ·0.094 kg · v²

2 · 4.7 J / 0.094 kg = v²

9.4 kg · m²/s² / 0.094 kg = v²

v = 10 m/s

The arrow will leave the bow with a velocity of 10 m/s.

5 0

4 years ago