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2 years ago

which type of front occurs when cold air and warm air are next to each other but are at a a stand still ?

Physics

2 answers:

vladimir1956 [14]2 years ago

7 0

Answer:

Stationary front

Explanation:

A stationary front is usually defined as a front that forms when the cold and warm air masses are next to each other but are at a standstill. This normally takes place when both the air masses push one another but neither of them is sufficiently strong to push the other. Both the air masses remain stationary because the winds blow parallel to the front instead of blowing perpendicular.

This type of front is basically responsible for the formation of cloudy weather that leads to rainfall or snowfall.

This stationary front is usually identified as the alternate red-colored semicircles and blue-colored triangles on weather maps. Both blue triangles and red semicircles are directed in the opposite direction.

algol132 years ago

3 0

A stationary front
hope this helps

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Why is it impossible to create a perpetual motion machine?

Tatiana [17]

Answer:

It is impossible to create a perpetual motion machine because some of the energy will always be lost in the conversion and therefore it will eventually stop.

Correct Answer : Option B

Explanation:

The perpetual motion machine is impossible machine as it is hypothetical working machine which would be in motion for an indefinite time in continuity of the motion. The continuity of motion for an indefinite time would mean that the working principle of the machine would never allow the dissipation of energy from the machine and all the machine would ultimately reserve all the energy and be converting it into forms without any loss.

This working principle is violation of first and second laws of thermodynamics and hence a machine will eventually lose some of its energy in every conversion of the working cycle and hence there will be a time where the machine would be stopped, and hence a perpetual motion machine cannot be made.

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2 years ago

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

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2 years ago

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Solar energy or sunlight...

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2 years ago

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When a car driving up a hill with constant speed: I. its kinetic energy is decreasing. II. its potential energy is constant. III

madam [21]

A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.

Understanding the concept

Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:

N is the normal reaction force acting in an upward direction

f_s is the static friction force exerted due to friction between the road and the tires of the car
f_k is the rolling friction force in the direction opposing that of the tire
mg is the force acting in a downward direction.

θ is the angle of inclination.

Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.

Learn more about motion on an incline here:

<u>brainly.com/question/13513083</u>

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1 year ago

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professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$