What happens to the density of gas when it compressed

When are tides highest? a. during the moon’s first quarter phase b. when the sun, Earth, and the moon are nearly in a line c. du

elena-14-01-66 [18.8K]

Ocean tides are highest when the sun, Earth, and the moon
are nearly in a line. That means at the times of New Moon
and Full Moon.

7 0

3 years ago

Read 2 more answers

Significant nomadic peoples who spoke a language derived from a single parent tongue were the

Blababa [14]

The Answer would be Indo - Europeans

7 0

2 years ago

A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: 

Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is:

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
W = 1250π ∫ x dx
W = 625π * (x2 – x1)

x2 = 14 ft + 15 ft = 29 ft

x1 = 14 ft + 1 ft = 15 ft

W = 625π * (29^2 - 15^2)
W = 385,000π ft-lbs = 1,209,513.17 ft-lbs

3 0

3 years ago

An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a In turn it forms an α-angle α

xxMikexx [17]

Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
$0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta - \alpha ) \frac{d}{u} - h = 0$
$4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0$
Hence obtain
$aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha$
The non-triial solution for X is
$X= \frac{b}{a}$

Answer:
$X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}$

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

Answer: X = 101.5 m

7 0

3 years ago

If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the in

poizon [28]

Answer:

The change in the internal energy of the gas 1,595 J

Explanation:

The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)

This is summarized in the expression:

ΔU= Q - W

where the heat absorbed and the work done by the system on the environment are considered positive.

Taking these considerations into account, in this case:

Q= 500 cal= 2,092 J (being 1 cal=4.184 J)

W=500 J

Replacing:

ΔU= 2,092 J - 500 J

ΔU= 1,592 J whose closest answer is 1,595 J

The change in the internal energy of the gas 1,595 J

6 0

3 years ago