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2 years ago

9

For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo

rce on their feet compared to when they are running in a straight line?

1 answer:

Snezhnost [94]2 years ago

5 0

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

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Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

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2 years ago

if the resistance of a car headlight is 15 ohm and the current through it is 0.60, what is the voltage across the headlight?

Strike441 [17]

Answer:

9 volts (assuming 0.60 is in Amperes)

Explanation:

Recall that Ohms law can be expressed as

V = IR, where

V = voltage,

I = current (given as 0.6. I'm going to assume that the units is Amperes because it is not given)

R = resistance (given as 15 ohm)

substituting the above values into the formula

V = IR

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3 years ago

Read 2 more answers

A mechanical wave starts when matter is disturbed by a source of ______.

Gwar [14]

Answer:

MATERIAL MEDIUM

Explanation:

Wave is a disturbance that travels through a medium and transfer energy from one point to another without causing any permanent displacement of the medium itself. The two forms of wave are the mechanical wave and the electromagnetic waves.

Mechanical wave is a wave with requires MATERIAL MEDIUM for its propagation. This means that before wave can be propagated at times, material medium is needed e.g a ripple tank. A ripple tank is a mechanical device that generates waves using an instrument called stroboscope attached to it. This kind of wave requires an external source before it can propagate compared to electromagnetic waves that does not require material medium for its propagation.

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2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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3 years ago

Volcanoes often form belts along plate boundaries. How does the piece of art illustrate that this statement holds true for North

STALIN [3.7K]

Answer:

The formation of volcanic arcs, trenches and subsidence leads to the production of volcanic eruptions.

Explanation:

North America plate makes a convergent boundary along with the pacific oceans. The is man reason for the development of the volcanoes and the origin of the earthquakes. The collusion of both these pales leads to the formation of a volcanic ring of fire that can be seen in the pacific ocean.

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2 years ago