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Shtirlitz
3 months 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:
ValentinkaMS [3.4K]3 months ago
7 0

Answer: 339.148N

Explanation:

Given data:

Time (t) = 47s

Initial speed (U) = 0m/s

Final speed (V) = 9.5m/s

Mass of B = 540kg

Frictional force on B = 230N

Since both boats are linked, movement of A causes B to move as well.

What is the acceleration of boat A?

Applying the motion formula:

V = u + at

9.5 = 0 + a * 47

a = 9.5 / 47

a = 0.2021 m/s²

To determine the force necessary to accelerate boat B, as both boats experience the same force:

F = Mass * acceleration

F = 540 * 0.2021 = 109.14N

Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:

Tension = frictional force + applied force = (109.14 + 230)N = 339.148N

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
4 months ago
Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and
Softa [3030]
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcos</span>β
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For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
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3 0
3 months ago
Read 2 more answers
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