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shusha [124]
3 years ago
12

A gnat takes off from one end of a pencil and flies around erratically for 41.641.6 s before landing on the other end of the sam

e pencil. If the gnat flew a total distance of 6.656.65 m, and the pencil is 0.04630.0463 m long, find the gnat's average speed and the magnitude of the gnat's average velocity.
Physics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

average speed  is 0.159 m/s

average velocity = 0.0011 m/s

Explanation:

given data

time = 41.6 s

total distance = 6.65 m

length = 0.0463 m

to find out

average speed and the magnitude average velocity

solution

we know average speed formula is

average speed = \frac{total distance}{total time}   ...............1

put here value

average speed = \frac{6.65}{41.6}

average speed  is 0.159 m/s

and

average velocity formula is

average velocity = \frac{displacement}{total time}   ...............2

here displacement is initial point to final point and here is 0.0463 m

put here value

average velocity = \frac{0.0463}{41.6}

average velocity = 0.0011 m/s

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Travka [436]

Hello!

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A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
GalinKa [24]

Answer:

a = 52s²

Explanation:

<u>How to find acceleration</u>

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

<u>Solve</u>

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

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7 0
3 years ago
What is the impulse
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Impulse is the integral of a force, F.
Hope this helps.
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5 0
3 years ago
An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
kvv77 [185]

Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

4 0
3 years ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
Bad White [126]
A) Agreed. 
<span>b) Value agreed but units should be W (watts). </span>

<span>c) Here's one method... </span>

<span>15 miles = 24140 m </span>

<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>

<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>

<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>

<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>

<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>

<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
8 0
3 years ago
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