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SpyIntel [72]
1 year ago
10

A speedboat with a mass of 531 kg (including the driver) is tethered to a fixed buoy by a strong 30.3 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 12900 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:
zaharov [31]1 year ago
4 0

As the tension in the rope is constant therefore the boat engine will also be pushing the boat with the same force as that of the drag force that is given to be 450 Newton.

<h3>What is a drag force?</h3>

Drag force is similar to friction force but is experienced by a moving body in a fluid. Direction of drag force is a positive to the direction of motion.

<h3>What is centripetal force?</h3>

Centripetal force is not a new force in nature but a manifestation of some force that causes rotation of a body.

Given:

Mass of the boat = 531 kg

Tension in the cable = 12900 Newton

Drag force experienced by the boat = 450 N

The boat along with the man is rotating using a cable which experiences a constant tension of 12900 Newton.

The boat's engine should also be generating a force perpendicular to the circular path travelled by the boat.

Now the tension in the string is constant this implies that all the force that the both engine is generating is being used in balancing the drag force exerted by water on the boat.

Therefore, the force produced by the engine will be equal to 450 Newton which will be used to balance the drag force.

Learn more about drag force here: brainly.com/question/27817330

#SPJ1

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A current of 3.6-A flows through a conductor. Calculate how much charge passed through and cross-section of the conductor in 25s
Anastaziya [24]

Answer:

90 C

Explanation:

Electric current: This can be defined as the rate of flow of electric charge in a circuit. This can be expressed mathematically as,

I = dQ/dt

dQ = Idt

∫dQ = ∫Idt

Q = It................................ Equation 1

Where Q = amount of charge, I = current, t = time.

Given: I = 3.6 A, t = 25 s.

Substituting into equation 1,

Q = 3.6(25)

Q = 90 C.

Hence the amount of charge passing through the cross section of the conductor = 90 C

6 0
3 years ago
Three point charges are fixed in place in a right triangle, as shown in the figure.
8090 [49]
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3 years ago
Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0
3 years ago
Can someone check my answers and tell me if their correct?
Otrada [13]

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read.  the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi =  2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 =  (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0  m/s   But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11


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Magnitude of the normal force exerted by en in the figure below. What is the
vekshin1

Answer:

B

Explanation:

Reason is that the suitcase is exerted downward and when it moves downward the equation is mgsin tita

6 0
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