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Sauron [17]
2 years ago
10

A rope connects boat A to boat B. Boat A starts from rest and accelerates to a speed of 9.5 m/s in a time t = 47 s. The mass of

boat B is 540 kg. Assuming a constant frictional force of 230 N acts on boat B, what is the magnitude of the tension in the rope that connects the boats during the time that boat A is accelerating?
Physics
1 answer:
Contact [7]2 years ago
7 0

Answer: 339.148N

Explanation:

Data

Time (t) = 47s

U = 0m/s

V = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Both boats are connected so if A moves, B moves too.

Acceleration of boat A =?

Using equation of motion,

V = u + at

9.5 = 0 + a*47

a = 9.5 / 47

a = 0.2021 m/s²

The force required to accelerate boat B since it's the same force moving both boats =?

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

A frictional force of 230N exists on boat B

Total force (Tension) = frictional force + normal force = (109.15 + 230)N = 339.148N

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Answer:

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Explanation:

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The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

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Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

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2 years ago
Read 2 more answers
A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl
Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  \vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s

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    The time considered is  t =  10 \  seconds

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 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

Here the negative sign in \frac{dx}{dt} is because of the negative sign that is attached to x in the equation

 So

       \frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]

From the given equation of velocity field

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So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

substituting the given values of x and t

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\frac{dT}{dt} = -0.8 +0.84 + 0.976  

\frac{dT}{dt} = 1.016 ^oC/m  

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I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
5 0
2 years ago
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