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Kaylis [27]
3 years ago
14

A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of

the of the motion of the yacht. By resolving the wind force parallel and perpendicular to the direction of motion, determine the component of the wind force in the direction of motion.

Physics
1 answer:
damaskus [11]3 years ago
4 0
The parallel component is given by
F=180cos(25)=163.14N
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A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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Centrifugal force is an inertial force that appears to act on all objects when viewed in a rotating frame of reference.

This force is directed away from the center around which the body is moving.

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

This is force that acts on a body moving in a circular path and is directed towards the center around which the body is moving.

While centripetal force is directed towards to the center, the centrifugal force is directed away.

Thus, the force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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3. A cart with mass of 30 kg is traveling with a velocity of 4.0 m/s. The
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39 m/s

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A curved line going up indicates the object is
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The Andromeda galaxy is the closest major galaxy to our own. Andromeda shows a distinct blue-shift of light when we analyze it.
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At the present time, the only way we know of that light can get shifted
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4 0
3 years ago
An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
MArishka [77]

Answer:

2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

800 = 4.9t²

t² = 163.26...

t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

7 0
3 years ago
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