The diagram shows the trajectory of a ball that

andreev551 [17]

For a 166-lb person, a dose of 452 milligrams is prescribed to treat asthma.

<h3>How would you define asthma?</h3>

Lung damage is a side effect of asthma. Repeated episodes of coughing, dyspnea, chest tightness, and wheezing are also brought on by it. By taking medication and avoiding the triggers that can set off an attack, asthma can be managed. Outdoor allergens, such as pollen from grass, trees, and weeds, are common asthma triggers. Different allergens and irritants might function as triggers for different people. Dust mites, cockroaches, mold, and pet dander are examples of indoor allergies. Air irritants like smoke, chemical fumes, and powerful scents.

Mass = 166 lb×(453.59 g/1 lb)×(1kg/1000g)=75.30 kg

Dose = 75.30 kg ×(6.00 mg/1 kg)=452 mg

Elixophyllin has a 452 mg dose for a 166 lb person with asthma.

To know more about asthma, click here:

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8 0

1 year ago

Explain this quote "ambition beats genius 99% of the time. "

klasskru [66]

Answer:

The idea that hard work is the most important aspect of new inventions existed before Edison gave his quote, however.

The idea behind this quote is that it is easy to have a good idea, or a creative insight. However, to follow through with that idea, and turn it into a reality, takes a level of patience and dedication that few people have.

Explanation:

7 0

3 years ago

Read 2 more answers

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$