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jok3333 [9.3K]
2 years ago
14

Barometers have long been used to predict the weather. Jay and Jeff were responsible for recording the class weather data each d

ay in March. One day, they noticed that there was a rapid change in the barometric pressure. The barometric pressure dropped down very, very quickly. What type of weather could Jay and Jeff expect soon? A) fog B) rain C) clearing D) drop in temperature
Physics
2 answers:
g100num [7]2 years ago
7 0

Answer:

B) rain

Explanation:

The barometer is a device that is used to measure atmospheric pressure and, consequently, the height at which someone rises, as well as to approximately predict atmospheric changes. Thus, a rising barometer indicates an increase in air pressure, whereas a falling barometer indicates a decrease in air pressure.

In environments with low atmospheric pressure and, therefore, heat, the heated air tends to rise and, when reaching higher altitudes, condenses, forms clouds and precipitates. In this situation, the barometric pressure drops very, very quickly, indicating that a rain is coming.

AfilCa [17]2 years ago
4 0

Answer:

B

Explanation:

You would have rain. You would have clouds, precipitation, and strong winds.

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The search for black holes involves searching for The search for black holes involves searching for Group of answer choices sing
trasher [3.6K]

Answer:

Large spherical regions from which no light is detected

Explanation:

A black hole is an object that has an extremely high density such that it possesses very powerful gravitational force that prevents the escape of all objects including light from it, and consumes nearby objects.

Due to the power of the gravitational force of a black hole, at the center, objects are infinitesimally compressed resulting in the inapplicability of the concept of space and time and the location is known as a singularity

Therefore, the search for black holes involves searching for <em>large spherical regions from which no light is detected</em>.

4 0
2 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
2 years ago
A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.
kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

E=mgh

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

K=\frac{1}{2}mv^2

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

8 0
3 years ago
A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0
2 years ago
How is the temperature of an object related to the average kinetic energy of its particle??
dusya [7]
By <span> It is directly proportional to the </span>average kinetic energy<span>.</span>
6 0
3 years ago
Read 2 more answers
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