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disa [49]
2 years ago
15

A stable air mass is most likely to have which characteristic? 1. turbulent air 2. poor surface visibility 3. showery precipitat

ion
Physics
1 answer:
Ne4ueva [31]2 years ago
3 0
A stable air mass is most likely to have POOR SURFACE VISIBILITY.
Stable air mass refers to those air mass that have marked stability in their lower layers. The characteristics of stable air mass include the following: cloud cover, smooth air, uninterrupted precipitation and low visibility.<span />
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Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

E_m = \frac{1}{2} m\omega_1^2A_1^2

The energy of the system having mass 2m is,

E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2

For the two expressions mentioned above remember that the variables mean

m = mass

\omega =Angular velocity

A = Amplitude

The energies of the two system are same then,

E_m = E_{2m}

\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2

\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}

Remember that

k = m\omega^2 \rightarrow \omega^2 = k/m

Replacing this value we have then

\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}

But the value of the mass was previously given, then

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
2 years ago
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
1 year ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar
Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s  

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s

Final momentum of the ball, p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s

Change in momentum of the ball, \Delta p=p_f-p_i

\Delta p=12-(-12)=24\ kg-m/s

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0
2 years ago
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