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3 years ago

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A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

the mule and pulls with his maximum force of 800 NN . The coefficients of friction between the mule and the ground are μsμsmu_s = 0.80 and μk=0.5μk=0.5.
Is the farmer able to move the mule?

1 answer:

jolli1 [7]3 years ago

4 0

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

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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

3 years ago

Find the current flowing out of the battery.

klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0

3 years ago

Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped

Akimi4 [234]

Answer: F = 102141N

Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

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

$a=\frac{11}{0.14}$

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>

6 0

2 years ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

$(X_{1},Y_{1})=(51,2390.7)$

Rewritting (1):

$Y=m(X-X_{1})+Y_{1}$ (2)

As Y is a function of X:

$Y=f_{(X)}=m(X-X_{1})+Y_{1}$ (3)

Substituting the known values:

$f_{(X)}=5.7(X-51)+2390.7$ (4)

$f_{(X)}=5.7X-290.7+2390.7$ (5)

$f_{(X)}=5.7X+2100$ (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

$f_{(81)}=5.7(81)+2100$ (7)

$f_{(81)}=2561.7$ (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0

3 years ago

According to Newton’s First Law of Motion, what will happen to a baseball thrown in outer space

Illusion [34]

Answer:

The baseball will stay in motion until another force act upon it.

Explanation:

3 0

3 years ago

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