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3 years ago

Average speed is the total distance divided by the

Physics

1 answer:

seraphim [82]3 years ago

8 0

<h2>Answer:</h2>

Time

<h2>Explanation:</h2>

The average speed of an object that is moving is defined as the distance traveled divided by the time of travel. You can measure the distance with a ruler and the time with a stopwatch. This can be expressed as the following formula:

$v=\frac{\Delta x}{\Delta t}$

For instance, if an object travels a distance $\Delta x=100m$ in 4 seconds, the the average speed is:

$v=\frac{100m}{4s} \\ \\ \therefore \boxed{v=25m/s}$

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A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3

Inga [223]

$m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.$

The momentul of the system preserves:

$m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.$

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with $E$ .

Now, we write the energy conservation law:

$\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E$

From the above equation, you find $E$ , and then conclude that the sound energy can certainly not be greater than this.

8 0

3 years ago

Can someone give me some tips of how to study. I have already tried writing everything in my lesson but it doesn't seem to work

larisa [96]

Highlighter, color coding sections that go together, flash cards, relaxing with someone while they read out questions ( this will help with testing anxiety too). If you get distracted easily use games to study or small sayings instead of the usual white and black sheet of paper.

3 0

3 years ago

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

3 years ago

How can you tell the difference between and erosion feature and a deposition feature?

Lunna [17]

Deposition moves that have already been eroded, and builds up sediment.
erosion is just the moving of sediment from one place to another

5 0

3 years ago

A puppy is running with a speed of 11.2 m/s towards it owner when it

Levart [38]

Answer:

-5.6m/s²

Explanation:

Given parameters:

Initial speed = 11.2m/s

Time of running = 2s

Unknown:

Acceleration of the puppy = ?

Solution:

Acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{Final speed - initial speed }{time}$

final speed = 0m/s

Acceleration = $\frac{0 - 11.2}{2}$ = -5.6m/s²

8 0

3 years ago