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Salsk061 [2.6K]
3 years ago
13

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

a while, and finally runs quickly back to her initial position along a straight line. Which of the following statements is true about the average speed and the magnitude of the average velocity of the student during her trip? Her average speed is greater than the magnitude of her average velocity. Her average speed is the same as the magnitude of her average velocity. Her average speed is less than the magnitude of her average velocity. The student's average speed and velocity cannot be compared without knowing the farthest point she reached.
Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

Answer:Average speed is greater than average velocity.

Explanation  :

Given

student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position

Let x be the length of track and the whole process takes t time

For average speed =\frac{distance\ traveled}{time\ taken}

Average speed=\frac{2x}{t}

For average velocity =\frac{Displacement}{time\ taken}

Since displacement is zero as she returns to its initial position.

Average velocity=0

Therefore Average speed is greater than average velocity.

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A 15.0 kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50 kg fish that is initially stationary. Ignore any drag effects of
Nana76 [90]

Answer

given.

Mass of big fish = 15 Kg

speed of big fish = 1.10 m/s

mass of the small fish = 4.50 Kg

speed of the fish after eating small fish =?

a) using conservation of momentum

m₁v₁ + m₂v₂ = (m₁+m₂) V

15 x 1.10 + 4.50 x 0 = (15 + 4.5)V

16.5 = 19.5 V

V = 0.846 m/s

b) Kinetic energy before collision

KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2

KE_1 = \dfrac{1}{2}\times 15 \times 1.1^2 + \dfrac{1}{2}m_2\times 0^2

KE₁ = 9.075 J

Kinetic energy after collision

KE_2= \dfrac{1}{2}(15+4.5)\times 0.846^2

KE₂ = 6.98 J

Change in KE = 6.98 - 9.075 = -2.096 J

hence,

mechanical energy was dissipated during this meal = -2.096 J

5 0
2 years ago
What force is needed to give a 0.25-kg arrow an acceleration of 196 m/s/s
kakasveta [241]
Force (f) = ?

Acceleration (a) = 196 m/s^2

Mass (m) = 0.25 kg

F = (m) • (a)

F = (0.25) • (196)

F = 49 N

Answer : 49 N

I hope that helps you!! Any more questions??

8 0
3 years ago
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The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

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Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

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A parallel circuit is a circuit in which the components are connected to a common junction. This implies that if one bulb goes out in a parallel connection, all the bulbs will go out.

As such, Jasmine need to correct the error by switching the headings on the columns adding the title parallel circuits.

Learn more about parallel connection: brainly.com/question/12400458

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2 years ago
What is the Sl unit for speed
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Answer: meter per second

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