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4 years ago

11

Suppose you were bungee jumping from a bridge while blowing a hand-held air horn. How would someone remaining on the bridge hear

the pitch of the air horn as time increased?
a. The pitch would get progressively lower (i.e., smaller frequency)

b. The pitch would get progressively higher (i.e., larger frequency)

c. The pitch would remain the same (i.e., constant frequency)

1 answer:

Readme [11.4K]4 years ago

6 0

Answer:

Answer to the question:

Explanation:

a. The pitch would get progressively lower (i.e., smaller frequency)

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What is the difference between mass and weight?

I am Lyosha [343]

Mass is the actual amount of material contained in a body and is measured by kg, gm, etc. Whereas weight is the force extorted by the gravity on that object mg. Note that mass is independent of everything but weight is different on earth, moon, etc.

5 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

(12 points) Analysis from the point where the block is released to the point where it reaches the maximum height i) Calculate th

ycow [4]

Answer:

i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁

ii) W = mg L sin θ , iii) Wₙ = 0

iv) W = - μ m g L cos θ x

Explanation:

With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.

The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp

At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane

i) Let's use Newton's second law

the reference system is with the x axis parallel to the ramp

Axis y

N - W cos θ = 0

X axis

T - W sin θ - fr = ma

the friction force is

fr = μ N

fr = μ mg cos θ

we substitute

T - m g sin sin θ - μ mg cos θ = m a

a = T / m - g (sin θ + μ cos θ)

With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.

T = 0

a₁ = -g (sin θ + μ cos θ)

v² = v₀² - 2a1 x

v = 0 at the highest point

x = v₀² / 2a₁

ii) the work of the gravitational force is

W = F .d

W = mg sin θ L

iii) the work of the normal force

the force has 90º with respect to the displacement so cos 90 = 0

Wₙ = 0

iv) friction force work

friction force always opposes displacement

W = - fr d

W = - μ m g cos θ L

4 0

3 years ago

Four boxes are sliding with constant speed before each box experiences and unbalanced force of 8 N. Which box would experience t

neonofarm [45]

Answer:

d

Explanation:

5 0

3 years ago

An electron of mass 9.11×10−31 kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to

dedylja [7]

Answer:

$F=1.509\times 10^{-16}N$

Explanation:

We have given that mass of electron $m=9.11\times 10^{-31}kg$

It is given that initial velocity u=0 m/sec

Final velocity v = $2.7\times 10^6m/sec$

Distance S = 2.2 cm = 0.022 m

According to third law of motion $v^2=u^2+2as$

$(2.7\times 10^6)^2=0^2+2\times a\times 0.022$

$a=165.68\times 10^{12}m/sec^2$

Accelerating force F is given by F = ma

So accelerating force $F=9.11\times 10^{-31}\times 165.68\times 10^{12}=1.509\times 10^{-16}N$

7 0

3 years ago