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Sholpan [36]
3 years ago
13

The motor of a four wheeler traveling along a muddy trail generates an average power of 7.48 104 W when moving at a constant spe

ed of 13 m/s. When pulling a log along the trail at the same speed, the engine must generate an average power of 8.50 104 W. What is the tension in the tow rope pulling the log?
Physics
2 answers:
Radda [10]3 years ago
7 0

Answer:

The tension in the tow rope pulling the log is 784.61 N

Explanation:

Given:

v = speed = 13 m/s

Pf = final power when the log is pulled = 8.5x10⁴W

Pi = average power = 7.48x10⁴W

The force required to move a car is equal to:

F_{i} =\frac{P_{i} }{v} =\frac{7.48x10^{4} }{13} =5753.85 N

F_{f} =\frac{P_{f} }{v} =\frac{8.5x10^{4} }{13} =6538.46 N

T = Ff - Fi = 6538.46 - 5753.85 = 784.61 N

DiKsa [7]3 years ago
7 0

Answer: 784.6 N

Explanation:

Given

Power of the motor, P1 = 7.48*10^4 W

Speed of the motor, v = 13 m/s

Power of the motor, P2 = 8.5*10^4 W

It is known that

Power = J/t, where

J = work, in joules

t = time, in seconds. Also,

Work = F * d, where

F = force

d = distance, thus on substituting, we have,

Power = Fd/t

Also, we know that

Velocity = d/t, where

d = distance,

t = time, thus, we could say that

Power = Force * Velocity

P = FV

To find the tension in the log, we use

F = P/V, where

P = P2 - P1

P = 8.5*10^4 - 7.48*10^4

P = 1.02*10^4 = 10200W

Force, F = 10200 / 13

F = 784.6 N

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sammy [17]

Answer:

Explanation:

I got the same thing. So, i don't know but good luck

3 0
1 year ago
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Taya2010 [7]

Answer:

  • 5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force  + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} ,  mass = density x volume

  • buoyant force =  density x volume x acceleration

        where density is the density of air for the buoyant force

        buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

  • force by helium =  density x volume x acceleration

        force by helium =  0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

  • force of its weight = mass of rubber x acceleration

        force of its weight = 0.0125 x 9.8 = 0.1225 N

  • Tension = buoyant force  + force by helium + force of weight of rubber

         the force  of weight of rubber and of helium act downwards, so they      

          carry a negative sign.

  • Tension = 6.54 - 0.917 - 0.1225 = 5.5 N
8 0
3 years ago
Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
tankabanditka [31]

Answer:

a) 24.33 m of distance.

b) 34.55° east of the south.

Explanation:

The question is incomplete. The whole exercise is the following:

<em>"Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 28.0 m in a direction 60.0° west of north. Jane walks 12.0 m in a direction 30.0° south of west. They then stop and turn to face each other. </em>

<em>(a) What is the distance between them? </em>

<em>(b) In what direction should Ricardo walk to go directly toward Jane?</em><em>"</em>

Now that we know what we need to do in this question, let's head for every part of the problem.

<u>a) Distance between Ricardo and Jane</u>

In this case, we need to analyze the given data:

Ricardo (which we will call R) is 28 m from the starting point at 60° west of north, and Jane (J) is 12 m at 30° south of west. So the distance between them, will be the point where they both stop and face each other. This point can be seen in the image attached (See picture).

Let's call the distance between them as "D", to get the distance of D, according to the picture will be:

D = J - R   (1)

However, as they are facing in different angles and directions, we cannot do the difference of their values distance just like that. In order to do that, we need to calculate the components in the "x" and "y" axis of each vector. In that way, we can get the components of x and y of the Distance D, and then, the whole distance between them will be:

D = √Dx² + Dy²     (2)

So, let's get the components of x and y of R and J.

For Ricardo (R):

Rx = R sin60° = 28 sin60° = -24.25 m

Ry = R cos60° = 28 cos60° = 14 m

The sign "-" it's because R it's on the second quadrant, therefore in x, we'll have to add the negative.

For Jane (J):

Jx = J cos30° = 12 cos30° = -10.39 m

Jy = J sin30° = 12 sin30° = -6 m

Again, the negative is added because J is on the third quadrant.

Now that we have the components, let's calculate vector D using expression (1):

Dx = -10.39 - (-24.25) = 13.86 m

Dy = -6 - 14 = -20 m

Now, using expression (2) we can finally know the distance between Jane And Ricardo:

D = √(-20)² + (13.86)²

<h2>D = 24.33 m</h2>

This is the distance between Jane and Ricardo.

b) Direction of Ricardo walking to Jane

In this case, we already have the components of x and y of the distance between them, so, to know the direction:

Tanα = Dy/Dx

α = tan⁻¹ (Dy/Dx)

Replacing the values we have:

α = tan⁻¹ (-20/13.86)

α = 55.45°

Which should south of east or:

β = 90 - 55.45

<h2>β = 34.55°</h2>

Ricardo should walk 34.55° east of south

Hope this helps

8 0
3 years ago
When the balloon sticks to the wall (assuming it sticks to the wall). It is
inn [45]
The answer is false
8 0
3 years ago
A two-slit pattern is viewed on a screen 1.26 m from the slits. If the two fourth-order maxima are 53.6 cm apart, what is the to
Anettt [7]

Answer:

= 6.55cm

Explanation:

Given that,

distance = 1.26 m

distance between  two fourth-order maxima = 53.6 cm

distance between central bright fringe and fourth order maxima

y = Y / 2

  =  53.6cm / 2

  = 26.8 cm

  =0.268 m

tan θ = y / d

         = 0.268 m /  1.26 m

         = 0.2127

       θ = 12°

4th maxima

d sinθ = 4λ

d / λ = 4 / sinθ

d / λ = 4 / sin 12°

d / λ = 19.239

for first (minimum)

d sinθ = λ / 2

sinθ =  λ / 2d

       =  1 / 2(19.239)

       = 1 / 38.478

       = 0.02599

    θ =  1.489°

tan θ = y / d

y = d tan θ

  = 1.26 tan 1.489°

  = 0.03275

the total width of the central bright fringe  

Y = 2y

  = 2(0.03275)

  = 0.0655m

  = 6.55cm

4 0
3 years ago
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