For a cars speedometer what is the calculated speed over an entire trip

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

2 years ago

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

2 years ago

IIIIIIIIII neeeeeed helppppp again?

Juliette [100K]

Answer:

The Principle of Progression

(I searched it up since I never learned this)

Explanation:

The principle of progression states that a person should start slowly and increase exercise gradually. Since Mandy is just getting started on her exercise routine, she should begin with a few workouts over a large span of time, then work her way up so she can do more workouts in a shorter span of time.

7 0

2 years ago

Astronomers are getting information about dark energy from

Thepotemich [5.8K]

The only way of telling about dark energy is our observation of how the universe has been expanding. It basically works the opposite as gravity, pushing things away from it. Thus, the closest answer would be D. The shape of galaxies in cluster galaxies.

6 0

3 years ago

Read 2 more answers

Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its

dolphi86 [110]

Answer: $4.487(10)^{11}m$

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

<em />

This law states a relation between the orbital period $T$ of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size $a$ of its orbit: