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Sindrei [870]
3 years ago
7

For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m

agnitude of the resistive force exerted by the water (in N) on the cruise ship?
Physics
1 answer:
Leviafan [203]3 years ago
4 0

Answer:

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

\Sigma F = F-R = 0 (Eq. 1)

Where:

F - Forward trust, measured in newtons.

R - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: (F = 6.85\times 10^{5}\,N)

R = F

R = 6.85\times 10^{5}\,N

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

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Magnetic domain structure is responsible for the magnetic behavior of ferromagnetic materials like iron, nickel, cobalt and their alloys, and ferrimagnetic materials like ferrite. ... Magnetic domains form in materials which have magnetic ordering; that is, their dipoles spontaneously align due to the exchange interaction.
4 0
3 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Vikki [24]

Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

              m = 146.28 kg = 146 kg

Mass of the cart = 146 kg

3 0
3 years ago
German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ???
fredd [130]

Answer:

\Delta x = 5.47 \times 10^{-9} m

Explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

\Delta x \times \Delta P = \frac{h}{4\pi}

so here we know that

\Delta v = 0.01 \times 10^6 m/s

m = 9.11 \times 10^{-31} kg

so we have

\Delta x \times (9.11 \times 10^{-31})(0.01 \times 10^6) = \frac{6.26 \times 10^{-34}}{4\pi}

\Delta x = 5.47 \times 10^{-9} m

5 0
3 years ago
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
2 years ago
Find the westward component of a resultant vector 85.42 unit, 23 degrees W of N
Sindrei [870]

Since the angle is West of North, therefore to find for the westward component (horizontal component) of the vector, we use the sin function:

sin θ = opposite side / hypotenuse = westward component / resultant vector

So the westward component (x) is:

x = 85.42 sin 23

<span>x = 33.38 unit</span>

5 0
3 years ago
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