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mezya [45]
2 years ago
13

A speedboat with a mass of 511 kg (including the driver) is tethered to a fixed buoy by a strong 31.5 m cable. The boat's owner

loves high speed, but does not really want to go anywhere. So the owner revs up the boat's engine, makes a lot of noise, and runs the boat in circles around the buoy with the cable supplying all the necessary centripetal force. When the tension of the cable is steady at 13300 N, with what force is the boat's engine pushing the boat? Different physics textbooks treat drag force somewhat differently and use different formulas. For the present purpose, take the water's drag force on the boat to be (450 kg/m)×2, where denotes the boat's speed. Ignore any drag force on the cable.
Physics
1 answer:
Lady_Fox [76]2 years ago
6 0

Answer:

its is 23.09 kb

Explanation:

i did this awsner

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This is a measure of force per unit area.
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If no work is done, the total energy is not changed. In that situation, if the kinetic energy changes, what must happen to the p
Norma-Jean [14]

Answer:

the potential energy will also change

Explanation:

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(b) If the object is at 330 feet and its instantaneous velocity is 3 feet per minute at 30 minutes, what is the approximate posi
ELEN [110]

Answer:

The final position is 36 feet.

Explanation:

initial position, d = 330 feet

speed, v = 3 feet per minute

time, t = 30 minute

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time interval = 2 minute

So, the distance in 2 minutes is

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8 0
3 years ago
A wire of length 6cm makes an angle of 20° with a 3 mT
Crazy boy [7]

Answer:

Approximately 7.3 \times 10^{-3}\; \rm A (approximately 7.3\; \rm mA) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let \theta denote the angle between the wire and the magnetic field.

Let B denote the magnitude of the magnetic field.

Let l denote the length of the wire.

Let I denote the current in this wire.

The magnetic force on the wire would be:

F = B \cdot l \cdot I \cdot \sin(\theta).

Because of the \sin(\theta) term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is 90^\circ.)

In this question:

  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
  • B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T.
  • l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m.
  • F = 1.5\times 10^{-4}\; \rm N.

Rearrange the equation F = l \cdot I \cdot \sin(\theta) to find an expression for I, the current in this wire.

\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}.

5 0
2 years ago
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