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dalvyx [7]
3 years ago
15

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

d the other an angle 11degrees east of north, as they pull the tanker a distance 0.83km toward the north. What is the total work they do on the supertanker?
Physics
1 answer:
sp2606 [1]3 years ago
7 0

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

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natima [27]
The magnitude of the force of gravity acting between the particles is:
F=G\frac{m_1.m_2}{d^2} 
The weight of each particle is:
P=mg
Now let's plug in the numbers knowing that G=6.67\times10^{-11} , g=9.81, d=0.8 and m1 and m2 are already given in kilograms. We get then:
P_1=m_1.g=78.48N
P_2=m_2.g=117.72N
F=G\frac{m_1.m_2}{d^2}=1.00\times10^{-8}N

This results shows us why we don't often see objects being attracted to each other, their mass is too small compared to the earth gravitational pull.

8 0
3 years ago
Which effect has been useful (and successful) in the search for and identification of black holes in the universe
kvasek [131]

Answer:

Detection of x-rays from a binary star undergoing mass exchange, where mass of component star can be determined.

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3 years ago
Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus
miss Akunina [59]

Answer:

(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2

Explanation:

Energy density in magnetic field is given as;

U_B = \frac{1}{2 \mu_o} B^2

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B is the magnetic field strength

Energy density of electric field

U_E = \frac{1}{2}\epsilon E^2

where;

E is electric field strength

Take the ratio of the two fields energy density

\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o}  *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2

Now replace E x L with IR

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2

Also, B = μ₀I / 2πr, substitute this value in the above equation

\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2

cancel out the current "I" and factor out μ₀

\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2

Finally, the equation becomes;

\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²

3 0
3 years ago
A cylinder measuring wide and high is filled with gas. The piston is pushed down with a steady force measured to be . Calculate
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Explanation:

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Now, radius of the cylinder will be as follows.

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              = \frac{2.6 cm}{2}

              = 1.3 cm

or,           = 0.013 m         (as 1 m = 100 cm)

As, area of cylinder = \pi \times r^{2}

                                = 3.414 \times (0.013 m)^{2}

                                = 5.77 \times 10^{-4} m^{2}

Relation between pressure and force is as follows.

             Pressure = \frac{Force}{Area}

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                         = 25996 N/m^{2}

Since, 1 N/m^{2} = 1 Pa           (as 1 kPa = 1000 Pa)

Therefore,   P = 25996 N/m^{2}  

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Thus, we can conclude that pressure of the gas inside the cylinder is 26 kPa.

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lesya [120]

Answer:

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Explanation:

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      p = h / λ

Where lam is called broglie wavelength

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This is the X-ray region

-bullet

     λ = 6.63 10⁻³⁴ / (1.90 10⁻³  313)

     λ = 1.11 10⁻³³ m

It is too small, only particle characteristics are observed

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Too small, only particle characteristics are visible

7 0
3 years ago
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