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3 years ago

Too much skepticism can

Physics

2 answers:

Schach [20]3 years ago

6 0

Too much skepticism will lead one to doubt everything while too little will lead to gullibility.

Gennadij [26K]3 years ago

4 0

Make you a hard head?

You might be interested in

What must crystalline solids have

liberstina [14]

Crystalline solids must have a specific, orderly arrangement of atoms to be considered so.

7 0

3 years ago

Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if

sergij07 [2.7K]

Explanation:

a) $I=\displaystyle \sum_{i}m_ir_i^2$

where $r_i$ is the distance of the mass $m_i$ from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

$I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2$

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

$\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2$

7 0

3 years ago

An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l

butalik [34]

Answer:

a) f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$ , b) Δf = 2 f₀ $\frac{v}{c}$

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

f ’= fo $\frac{c+v}{c}$

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

f ’’ = f’ $\frac{c}{ c-v}$

where c represents the sound velocity in stationary blood

therefore the received frequency is

f ’’ = f₀ $\frac{c}{c-v}$

let's simplify the expression

f ’’ = f₀ \frac{c+v}{c-v}

f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$

b) At the low speed limit v <c, we can expand the quantity

(1 -x)ⁿ = 1 - x + n (n-1) x² + ...

$( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}$

f ’’ = fo $( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )$

f ’’ = fo $( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )$

leave the linear term

f ’’ = f₀ + f₀ 2 $\frac{v}{c}$

the sound difference

f ’’ -f₀ = 2f₀ v/c

Δf = 2 f₀ $\frac{v}{c}$

4 0

2 years ago

What is the difference in the pitch when you are in front of the car compared to behind the car?

IceJOKER [234]

Answer:The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener.

As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound. The opposite happens when the source of sound waves moves away from the listener.

Explanation:

7 0

3 years ago

As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s2 as it slows from 9.0 m/s

Dennis_Churaev [7]

Answer:

The time interval of acceleration for the bus is 2.20 seconds

Explanation:

Acceleration is the rate of change of velocity

→ $a=\frac{v-u}{t}$

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

The given is:

The uniform acceleration = -4.1 m/s²

The bus slows from 9 m/s to 0 m/s

We need to find the time interval of acceleration for the bus

Lets use the rule above

→ a = -4.1 m/s² , v = 0 m/s , u = 9 m/s

→ $-4.1=\frac{0-9}{t}$

Multiply both sides by t

→ -4.1 t = -9

Divide both sides by -4.1

∴ t = 2.20 seconds

<em>The time interval of acceleration for the bus is 2.20 seconds</em>

3 0

3 years ago