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3 years ago

A man pushes a lawn mower on a level lawn with a force of 186.5 N. If 3.8% of this

Physics

1 answer:

Paladinen [302]3 years ago

4 0

Explanation:

The force is 186.5 N. 3.8% of the force is directed downward. So the vertical component of the force is:

Fᵧ = 0.038 (186.5 N)

Fᵧ = 7.087 N

The horizontal component is found with Pythagorean theorem:

F² = Fₓ² + Fᵧ²

(186.5 N)² = Fₓ² + (7.087 N)²

Fₓ = 186.4 N

So the work done is:

W = Fd

W = (186.4 N) (7.4 m)

W = 1379 J

Round as needed.

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Amanda selects a sample of college students. She obtains the students' scores on several personality scales, as well as on a tes

andreyandreev [35.5K]

Answer:

descriptive

Explanation:

In this scenario, Amanda would use descriptive statistics in order to present the raw data. That is because, this type of statistic fulfills her goal of summarizing the raw data, while still providing clear and accurate quantitative analysis regarding the different features of the study through the different tests done. Including the personality scales and creativity test data. Since this provides such information it would best fit the goal of what Amanda is trying to accomplish.

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3 years ago

What are the four steps in the machine cycle?

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3 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

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3 years ago

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to 6°C at an average

Ratling [72]

Answer:

The correct option is;

Explanation:

Here we have the Coefficient Of Performance, COP given by

$COP = \frac{Q_{cold}}{W} = 3.1$

The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have

$Q_{cold}$ = m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J

Therefore, plugging in the value for $Q_{cold}$ in the COP equation we get;

$COP = \frac{197.39 }{W} = 3.1$ which gives

$W = \frac{197.39 }{3.1} = 63.674 \ J$

Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.

The power input to the refrigerator is approximately 64 W.

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3 years ago

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