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2 years ago

List three physical properties of a soft drink

Physics

2 answers:

Eva8 [605]2 years ago

8 0

density, amount of gas dissolved, refractive index

topjm [15]2 years ago

4 0

Examples of physical properties include: color, shape, size, density, melting point, and boiling point.

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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

4 0

3 years ago

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

2 years ago

A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w

Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

$h_1=\frac{L}{2}sin\theta$

The center of mass of the ladder ends at a height of:

$h_2= \frac{L}{2}sin90$ =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = $mg(h_2-h_1)$

now $h_2-h_1= 1-sin\theta$

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0

3 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

Rainbow [258]

An object in motion will stay in motion, therefore the person will still be going the same speed as the car was going before the collision

6 0

3 years ago

Please help with these i dont know how to do them

Svetradugi [14.3K]

20.0M/S. IS THE HORIZONTAL VELOCITY OF THE BALL JUST BEFORE IT REACHES THE GROUND

12.2 SECONDS IS THE APPROXIMATE TOTAL TIME REQUIRED FOR THE BALL

3 0

3 years ago