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Katyanochek1 [597]
3 years ago
15

Flow around curved height contours requires the incorporation of the centrifugal force. What is the general term to describe the

winds that flow along a curved trajectory above the level where friction plays a role?
Physics
1 answer:
lilavasa [31]3 years ago
4 0

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.

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Which of the following is an example of an even population distribution?
icang [17]
Even population distribution refers to a type of population distribution in which the arrangement is done in such a way that the distance between neighboring individual is maximized and uniform. This type of population distribution is usually found on the farm land, where the space in which the crops are planted have been carefully measured out. Thus an example of even population distribution is corn planted in a field.
8 0
3 years ago
A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient
Basile [38]

Answer:

v_o = 4.54 m/s  

Explanation:

<u>Knowns  </u>

From equation, the work done on an object by a constant force F is given by:  

W = (F cos Ф)S                                   (1)  

Where S is the displacement and Ф is the angle between the force and the displacement.  

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:  

K.E=1/2m*v^2                                       (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:  

W = K.E_f-K.E_o                                 (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020  

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:  

∑F_y=N-mg

     N=mg

Thus, the kinetic friction force is:  

f_k = μ_k*N  

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.  

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:  

W_f=(f_k*cos(180) s

      =-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:  

W= -K.E_o    

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force  

W_f= -K.E_o  

From equation (2), the work done by the friction force in terms of the initial speed is:  

W_f=-1/2m*v^2  

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:  

-μ_k*mg*s = -1/2m*v^2  

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:  

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s  

v_o = 4.54 m/s  

6 0
3 years ago
Read 2 more answers
Which statement below concerning the photoelectric effect is true? View Available Hint(s) Which statement below concerning the p
kakasveta [241]
There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.
8 0
3 years ago
What is measured using micrometer?<br><br>​
Fantom [35]

Explanation:

A micrometer, sometimes known as a micrometer screw gauge, is a device incorporating a calibrated screw widely used for accurate measurement of components in mechanical engineering and machining as well as most mechanical trades, along with other metrological instruments such as dial, vernier, and digital calipers

6 0
2 years ago
Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in
algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, F=2\times 10^4\ N

The impulse delivered to the object is given by the change in momentum as :

F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0
3 years ago
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