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lord [1]
3 years ago
15

How are steam powered boats different from sailboats

Physics
2 answers:
lord [1]3 years ago
6 0
Hi there!

Steam powered boats use steam to propel and move the boat, while sailboats rely in the angle of the sail and the wind to move and turn.

Hope this helps!
Alex3 years ago
3 0
Steam powered boats run from an engine, using power, gears, and cranks to function.

Sailboats use no engine and rely on winds to move and change the direction of the boat. 
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Flow around curved height contours requires the incorporation of the centrifugal force. What is the general term to describe the
lilavasa [31]

Answer: Gradient Wind

Explanation:

Gradient wind, is the wind that accounts for air flow along a curved trajectory. It is an extension of the concept of geostrophic wind; for example the wind assumed to move along straight and parallel isobars (lines of equal pressure). The gradient wind represents the actual wind better than the geostrophic wind, especially when both wind speed and trajectory curvature are large, because they are in hurricanes and jet streams.

4 0
2 years ago
If the AMA of the inclined plane below is 2, calculate the IMA and efficiency. IMA = Efficiency =
wlad13 [49]

Answer:

IMA = 2.5 metres

EFFICIENCY = 80%

Explanation:

The AMA of a machine is referred to as the Actual Mechanical Advantage of a machine, calculated as the ratio of the output to the input force.

The Ideal Mechanical Advantage is the ratio of the input distance to the output distance.

From the diagram, the input distance which is also the distance moved by effort  = 5metres

The load distance (output distance) = 2 metres

IMA = INPUT DISTANCE / OUTPUT DISTANCE

IMA = 5metres / 2 metres = 2.5 meters

Efficiency is the ratio of AMA TO IMA

AMA = 2, IMA = 2.5

EFFICIENCY = AMA / IMA

EFFICIENCY = (2 / 2.5) × 100%= 0.8 × 100%

EFFICIENCY = 80%

5 0
2 years ago
A 5 kg ball is suspended on one cable. Calculate the tension in the cable
ludmilkaskok [199]

Answer:

49N

Explanation:

F=ma

We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2

F=5kg*9.8m/s^2

 = 49N

Hope this helps!

7 0
2 years ago
A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight
Ede4ka [16]

it will take 36 seconds


8 0
2 years ago
1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0
2 years ago
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