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3 years ago

If the runner completes exactly one lap what is her average speed? her average velocity? Explain your answer.

1 answer:

4 0

Average speed = (total distance covered) / (time to cover the distance).

Runner's average speed = (length of 1 lap) / (her time to run 1 lap).

________________________________

Average velocity =

(straight-line distance between start-point and end-point)
divided by
(time to go from start-point to end-point).

The runner's start-point and end-point are the same place.
The distance between them (her 'displacement') is zero.
Her average velocity = zero .

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An aquarium 8 m long, 4 m wide, and 4 m deep is full of water. The hydrostatic pressure on the bottom of the aquarium is ____ Ne

MA_775_DIABLO [31]

Answer:

Explanation:

Given

length=8 m

width=4 m

deep=4 m

Area of base $A=4\times 8=32 m^2$

hydro static Pressure on bottom $=\rho gh$

$P=10^3\times 9.8\times 4=39.2 kPa$

Force on bottom $=P\times A=1254.4 KN$

Now pressure at the side wall

$F=\rho gA \bar{x}$

where $\bar{x}$ is the mean length of wall = 4 m

$F=10^3\times 9.8\times 4\times 8\times 2=627.2 KN$

4 0

3 years ago

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive const

Anvisha [2.4K]

Answer:

The position of stable equilibrium is -a

And the period of small oscillations must be: c/(ma^3)

Explanation:

Since the potential is:

$V(x) = \frac{c x}{a^2+x^2}$

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

$V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}$

Which vanish for

x = a ; x =-a

The second derivative of V(x) is:

$V''(x) = \frac{8 c x^3}{\left(a^2+x^2\right)^3}-\frac{6 c x}{\left(a^2+x^2\right)^2}$

And:

$V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}\\$

Therefore:

The position of stable equilibrium is -a

And the period of small oscillations must be:

$\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}$

(c/(ma^3))^1/2

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c

7 0

3 years ago

Calculate the weight of the following:<br> a. A car with a mass of 2500kg

Alex Ar [27]

Answer:

24525 N

Explanation:

The <em>mass</em> of an object is a fundamental property of the object.

The <em>weight</em> of an object is the force of gravity on the object.

This is defined as follows

Weight (W)= mass(m) × weight of a unit mass(g)

( gravitational field intensity = g)

Assuming g= 9.81 ms⁻²,

W = mg

= 2500×9.81 = 24525 N

6 0

3 years ago

What are the three main classes of rock?

Vladimir [108]

Answer:

There are three kinds of rock: igneous, sedimentary, and metamorphic.

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3 years ago

What is the path of an electron moving perpendicular to a uniform magnetic field?

agasfer [191]

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Explanation:

this is your answer I hope it is helpful please mark me brainly

7 0

3 years ago