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3 years ago

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Dana is on a train traveling at a speed of 20 km/h. Dana walks from the front of the train to the back of the train at a speed o

f 4 km/h. What is Dana's speed relative to the ground?

1 answer:

dezoksy [38]3 years ago

5 0

Answer:

Diana's speed relative to ground is <u>16 km/h</u> in the direction of motion of train.

Explanation:

Given:

Velocity of train in forward direction is, $v_{t,g}=20\ km/h$

Here, $v_{t,g}\to \textrm{velocity of the train relative to ground.}$

Velocity of Diana relative to train in the backward direction is, $v_{D,t}=-4\ km/h$

Negative sign implies backward motion or motion opposite to the direction of train's motion. Here, Diana is walking from front of train to back. So, Diana is moving in the opposite direction.

Now, we know that, for two bodies 'A' and 'B', velocity of 'A' relative to ground is given as:

$v_{A,g}=v_{B,g}+v_{A,B}$

Therefore, velocity of Diana relative to ground is given as:

$v_{D,g}=v_{t,g}+v_{D,t}$

$v_{D,g}=(20-4)\ km/h$

$v_{D,g}=16\ km/h$

So, Diana's speed relative to ground is 16 km/h in the direction of motion of train.

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You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

<em> 11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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3 years ago

Quel est le type de rayonnement produit par la radioactivité

Dmitriy789 [7]

Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:

-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons

-- "rayons bêta" ou "particules bêta" . . . flux d'électrons

-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée

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3 years ago

In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$

or

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s$

The time taken is 2.73414 seconds

The potential energy is given by

$U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J$

The change in potential energy is 467622.66798 J

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3 years ago

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tensa zangetsu [6.8K]

I’m pretty sure it’s c sorry if I’m wrong

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3 years ago

If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be

ale4655 [162]

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

The energy at rest which is the potential energy is 500J
This energy will be converted to kinetic energy in total after the arrow has been released.
This way, no energy is lost and we can account for the energy transformations occurring.

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3 years ago