If you run at 10 m/s for 1 minute, how far will you go?

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:

brainly.com/question/5009040

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8 0

1 year ago

An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0

3 years ago

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy.

statuscvo [17]

Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.

According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J

Minus sign indicates that work is done by the system.

7 0

3 years ago

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A hair dryer draws a current of 10 A plugged into a 120 V outlet. What is the resistance of the hair dryer?

Ganezh [65]

<span>This problem can be solved by the formula used to find resistance. The formula is R=V/I which basically means divide the Voltage by the Current to find the Resistance in an object. Ohm's law.</span>

8 0

3 years ago

Read 2 more answers

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago