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7 days ago

The city is thinking about starting a program to add a new chemical to the public pool. Before they do, they want to find out if

there are any health risks associated with this type of chemical. Which group would be the best to advise them on this topic?
a group of concerned citizens who oppose the program
the company that manufactures the chemical
the local pest pool companies who would be hired to take care of the pool
the government agency that regulates these types of chemicals

Physics

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A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w

ValentinkaMS [3465]

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h = v cos θ

v_h = 27 × cos 45°

= 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

= 74.375 m

The ball was in the air for 3.896 seconds

8 0

2 months ago

In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c

Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

$GDP=\pi } \frac{d^{2} }{4}$

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

$GDP_{b} =\pi \frac{3^{2} }{4}$

Conversely, in darkness, the diameter typically enlarges to 7 mm.

$GDP_{b} =\pi \frac{7^{2} }{4}$

This indicates an increase in light-gathering capacity.

$Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}$

Thus, the amount of light the eye can capture is 31.4 mm².

3 0

2 months ago

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

Yuliya22 [3333]

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x $\frac{1}{3}$ x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

= 2π x 11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s

= 128.23 x 10⁵ kgm² s⁻¹.

4 0

3 months ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Yuliya22 [3333]

Answer:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

Explanation:

In this scenario, we determine the initial velocity as follows:

$v_i = 7 \frac{m}{s}$

The final velocity in this instance can be expressed as:

$v_f = 13 \frac{m}{s}$

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

$v_f = v_i +at$

Solved for acceleration, we find:

$a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}$

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

$v_f = v_ i +a t$