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3 years ago

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You drive a car 2.0 km to the gas station, 4.0 km to the shopping mall to let your mother off, and 4.5 km back home. The trip ta

kes 20.0 min. What is your average speed?

1 answer:

slamgirl [31]3 years ago

8 0

60 i believe (ssssppppaaaccceeee)

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Answer: At several different levels, scientific awareness will enhance the quality of life, from the routine functioning of our daily lives to global issues. In energy, conservation, agriculture, health, transportation, communication, security, economics, recreation, and discovery, science informs public policy and personal decisions.

Explanation:

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Answer the attached question.<br>No Spam!<br>⚔️⚔️⚔️

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<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

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<h3> =1 m/s</h3>

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

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<h3> = 0</h3>

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

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<h3> = -1/3 m/s</h3>

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

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<h3> = 400 m</h3>

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3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

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