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3 years ago

Write the equality and conversion factors for the relationship between miles and hours for a car traveling at 55 mi/h

Physics

1 answer:

AysviL [449]3 years ago

5 0

Using Raoult’s Law, we can calculate for the vapor pressure using the formula:

P = x * P(pure)

Where x is the mole fraction of water, P(pure) = pressure of pure water, P = pressure when in solution

calculating for mole fraction of water:

x = (0.42kg / 18kg/kmol) / ((0.42kg / 18kg/kmol) + (0.075kg / 192.12kg/kmol))

x = 0.9835

Therefore:

P = 0.9835* P(71.93 mmHg)

P = 70.75 mmHg

You might be interested in

A ball rolls forward in the grass slowing down as it rolls?

olga nikolaevna [1]

Yes it does, uh huh. It slows down as it rolls. That's a fact.

In order for the ball to roll forward, it has to push grass out of the way. That takes energy. To bend each blade of grass out of its way, the ball has to use a tiny bit of the kinetic energy that it has, so it gradually runs out of kinetic energy. When its kinetic energy is all gone, it stops moving.

3 0

3 years ago

the measure of each exterior angle of a regular pentagon is ___ the measure of each exterior angle of a regular nonagon

Cerrena [4.2K]

Answer:

(a) 72°

(b) 40°

Explanation:

PENTAGON

First, we calculate the total angles in a Pentagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 5.

Hence, the total angle in a polygon is

180(5 - 2) = 180 * 3 = 540°

Therefore, each angle will be:

540°/5 = 108°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular pentagon will be:

180 - 108 = 72°

The exterior angle of a regular Pentagon is 72°

NONAGON

First, we calculate the total angles in a Nonagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 9.

Hence, the total angle in a polygon is

180(9 - 2) = 180 * 7 = 1260°

Therefore, each angle will be:

1260°/9 = 140°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular nonagon will be:

180 - 140 = 40°

The exterior angle of a regular Nonagon is 40°

4 0

3 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

3 0

3 years ago

Use the information to answer the following question.

Nostrana [21]

Answer:

The answer is B. :)

Explanation:

5 0

3 years ago

A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he

RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2} (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

4 0

3 years ago