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3 years ago

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Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is

the velocity of runner 1? A.17 m/s east. B.3 m/s east. C.17 m/s west. D.3 m/s west

1 answer:

Vsevolod [243]3 years ago

5 0

Answer:

<em>17 m/s west</em>

Explanation:

Runner 1 has velocity = 10 m/s west

runner 2 has velocity = 7 m/s east

From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of

velocity = 10 m/s + 7 m/s = <em>17 m/s west</em>

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A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

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3 years ago

A pendulum with a length of 1 meter is released from an initial angle of 15.0° after 1000s its amplitude has been reduced by fri

Setler [38]

Answer:

0.366×10^{-3} / s

Explanation:

θ = θmax e^{-bt/2m}

Given that

θ = 5.50°

θmax = 15.0°

So that we have

ln (θ / θmax) = -bt /2m

= - ln(5.50°/ 15.0°) / 1000s = b /2m

= b / 2m = 0.366×10^{-3} / s

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3 years ago

how long does it take sound to travel the distance between the two microphones? Given.:wave 1 of microphone 1 has T=2 sec and f=

Maksim231197 [3]

Answer:

0.00583 seconds

Explanation:

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A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget

avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

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olchik [2.2K]

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Explanation:

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