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shepuryov [24]
3 years ago
14

6. Regarding the formula Al2O3 which of the following is accurate? A. The coefficient 3 indicates that there are a total of thre

e atoms of oxygen present in the substance. B. The subscript 2 indicates that two atoms of oxygen are present in the substance. C. The chemical symbol Al indicates that oxygen is present in the substance. D. The subscript 2 indicates that two atoms of aluminum are present in the substance.
Physics
1 answer:
Norma-Jean [14]3 years ago
5 0
The correct option is D.
The chemical formula AI2O3 given in the question above gives the following information:
1. The subscript 2 indicates that two atoms of aluminium are present in the compound.
2. The subscript 3 indicates that three atoms of oxygen are present in the chemical compound.
3. The chemical symbol Al indicate that aluminium is present in the compound.
4. The chemical symbol O indicates that oxygen is present in the compound.
5.The subscript 2 and 3 show the ratio in which aluminium and oxygen atoms combine together.<span />
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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
Witch planet takes 84 earth year's to orbit the sun just once
alekssr [168]
Uranus takes 84 earth years to make a full rotation around the sun<span />
8 0
3 years ago
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Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se
RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy  = 1 / 7

λy / λx = 7 / 1

7 0
3 years ago
Which statements describe a situation with a displacement of zero? Check all that apply.traveling south for 30 miles, then turni
cluponka [151]

Answer:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

Explanation:

Displacement simply means the.change in position of an object. In a situation whereby the initial and final position are thesame, the displacement will be zero.

The statements that describe a situation with a displacement of zero include:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

5 0
3 years ago
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
1 year ago
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