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3 years ago

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You are running on a perfectly circular track that has a radius of 102 meters. What is the largest displacement you will ever at

tain?

1 answer:

tiny-mole [99]3 years ago

8 0

The largest possible displacement on a circular track is the straight-line distance between the starting point and the point directly opposite it, half-way around the circle. That's the diameter of the track ... 204 meters.

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

What is the resultant of a pair of one pound forces at right angles to each other?

OleMash [197]

Answer:

$F_R=\sqrt{2} \ pound.force$

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

$F_R=\sqrt{F_1^2+F_2^2}$

where:

$F_R=$ resultant force

$F_1\ \&\ F_2$ be the two of the forces to be added.

$F_R=\sqrt{1^2+1^2}$

$F_R=\sqrt{2} \ pound.force$

8 0

4 years ago

Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.

Yakvenalex [24]

Answer:

<h2>3.0 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{6}{2} \\$

We have the final answer as

<h3>3.0 m/s²</h3>

Hope this helps you

4 0

3 years ago

A person with a mass of 40 kg is sitting on a box. What is the value of the normal force

VMariaS [17]

normal force=mass*gravitational force

normal force=40*0

normal force=40

8 0

3 years ago

Which two quantities can be used to describe motion? A: Displacement and weight. B: speed and acceleration. C: speed and mass. D

HACTEHA [7]

The answer is B as all the other options contain quantities not related to describing motion

3 0

3 years ago