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yan [13]
3 years ago
13

The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approa

ching the dock at the moment when there is 10 feet of rope still left to pull in

Physics
1 answer:
Oduvanchick [21]3 years ago
7 0

Answer:

The boat is approaching the dock at a rate of <u>2.5 ft/s</u>.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, \frac{dl}{dt}=2\ ft/s

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)

Now, differentiating the above equation with time 't', we get:

2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

b^2=10^2-36\\\\b=\sqrt{64}=8\ ft

Now, substituting all the given values in equation (2) and solve for \frac{db}{dt}. This gives,

\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

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Answer:

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Explanation:

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3 years ago
Hi guys! pls help I asked this question almost 2 times and still didn't receive my answers.....Thanks in advance..
rjkz [21]

Answer:

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3 0
3 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0
3 years ago
Read 2 more answers
A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.
Kazeer [188]

Answer:

(a) Magnetic moment will be 17.212\times 10^{-4}A-m^2

(b) Torque will be 6.024\times 10^{-4}N-m

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2

Current is given as i=15mA=15\times 10^{-3}A

Number of turns N = 25

(A) We know that magnetic moment is given by magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by \tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m

4 0
3 years ago
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Answer:

B) with 9/10 submerged

Explanation:

m = mass of ice cube

\rho = density of soft drink

V = Volume of soft drink displaced

ice cube floats in the soft drink when the force of buoyancy on it balances its weight. Force of buoyancy acting on the cube in upward direction is same as the weight of the soft drink displaced. hence we can write

weight of ice cube = weight of soft drink displaced

mg = \rho V g

m = \rho V

we see that the acceleration due to gravity cancel out both side and hence it does affect as astronaut is on earth on in a lunar module.

8 0
3 years ago
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