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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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I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app
Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

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  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

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Learn more here:brainly.com/question/17280180

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3 years ago
A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe
Mariulka [41]
<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

  • A certain circuit is composed of two series resistors
  • The total resistance is 10 ohms
  • One of the resistor is 4 ohms

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • We have to find the value of other resistor?

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

We know that,

In series combination,

  • When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

<u>That </u><u>is</u><u>, </u>

Rn in series = R1 + R2 + R3.....So on

<u>Therefore</u><u>, </u>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω

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