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Sindrei [870]
4 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]4 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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By the way, voltage doesn't "run through" anything. Current does. That would be the 14 Amps.
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what is required to bring about a phase change 1.an increase or decrease in pressure 2.an increase or decrease in energy 3.an in
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Answer

Correct answers are 1.an increase or decrease in pressure 2.an increase or decrease in energy

Explanation

All existing matter can undergo phase change it means they may transform from one state to another. phase change of a matter  may occur due to change in energy and change in pressures.

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In some cases the matter doesn't want to undergo phase transformation. For example, oxygen will solidify at -361.8 degrees Fahrenheit at standard pressure.But , it can change to solid state at warmer temperatures when the pressure is increased.


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3 years ago
Read 2 more answers
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
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4 years ago
1pt A cannon fires a 5-kg ball horizontally from a
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Answer: Both cannonballs will hit the ground at the same time.

Explanation:

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a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

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and:

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And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

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