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Rufina [12.5K]
3 years ago
14

What are some modern uses for infrasound example, ultrasound can be used in the medical field to look inside people. Thanks!

Physics
1 answer:
lyudmila [28]3 years ago
5 0
Infrasound sound is any sound that are below the sound a human can hear, elephants use it to communicate with other elephants.
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An object in free fall travels a distance s that is directly proportional to the square of the time t. If an object falls 1088 f
ikadub [295]

Answer:

1,700feet

Explanation:

If an object in free fall travels a distance s that is directly proportional to the square of the time t, this can be represented mathematically as;

S = kt²where;

k is the proportionality constant

K = s/t²

s1/t1²= s2/t2²= Sn/tn²= k for values of the distance and time. Using the formula

s1/t1² = s2/t2² where;

s1 is the falling distance in time t1 s2 is the falling distance in time t2

Given s1 = 1088feet, t1 = 8secs, s2 = ? t2 = 10secs

Substituting this value in the formula to get s2, we have;

1088/8²= s2/10²

64s2= 108800

s2 = 108800/64

s2 = 1,700feet

This means the object will fall a distance of 1,700feet in 10seconds

5 0
3 years ago
Leverrier predicted that an invisible planet was pulling the planet Uranus off its predicted course around the sun. Likewise, hu
Alexandra [31]

Human beings are pulled off course as a result of the invisible forces of the <u>unconscious.</u>

According to Leverrier, he stated that an invisible planet was pulling the planet Uranus off its predicted course around the sun.

In such a way, human beings are pulled off course by the invisible forces of their unconscious minds. The unconscious minds of people control the thoughts of people.

Read related link on:

brainly.com/question/25588203

3 0
3 years ago
Whats an astral projection
Dominik [7]

Answer:Esotericism

Explanation:

it’s something that’s in intentional out of body experience

5 0
3 years ago
Read 2 more answers
the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300
Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

= 8,483.4 J

Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

4 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
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