1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nataly_w [17]
3 years ago
9

Winds are deflected to the (right/left) ___ as they move into a low pressure area in the Northern Hemisphere.

Physics
1 answer:
SashulF [63]3 years ago
8 0

Winds are deflected to the right as they move into a low pressure area in the Northern Hemisphere.

<u>Explanation:</u>

Winds decide the motion of ocean currents which forms the surface waves in the Earth's atmosphere to maintain the pressure region. The motion of ocean currents is based on Coriolis force which states the direction of motion of an object in a rotating system.

In the case of Earth, the Coriolis force has an effect on the ocean currents which are deflected from maximum to minimum pressure region in a curved path. So the winds formed by the ocean currents will generally get deflected at the right as they move into a low pressure area at the Northern Hemisphere from the high pressure region.

You might be interested in
Select the correct statement to describe when a sample of liquid water vaporizers into water vapor
AnnyKZ [126]
From what I know; When a sample of liquid water vaporizes into water vapor, the electrons in the water sped up due to heat. 
4 0
4 years ago
PLEASE HELP
Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 0
3 years ago
Which letter represents the wavelength of
Morgarella [4.7K]
The letter D represents the wavelength
4 0
3 years ago
Read 2 more answers
The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre
frozen [14]

Answer:

a) P_m=1.78\ Pa

b) f=79.5775\ Hz

c) \lambda=7.076\ m

d) v=563.06\ m.s^{-1}

Explanation:

<u>Given equation of pressure variation:</u>

\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]

We have the standard equation of periodic oscillations:

\Delta P=P_m\ sin\ (kx-\omega.t)

<em>By comparing, we deduce:</em>

(a)

amplitude:

P_m=1.78\ Pa

(b)

angular frequency:

\omega=2\pi.f

2\pi.f=500

∴Frequency of oscillations:

f=\frac{500}{2\pi}

f=79.5775\ Hz

(c)

wavelength is given by:

\lambda=\frac{2\pi}{k}

\lambda=\frac{2\pi}{0.888}

\lambda=7.076\ m

(d)

Speed of the wave is gives by:

v=\frac{\omega}{k}

v=\frac{500}{0.888}

v=563.06\ m.s^{-1}

8 0
3 years ago
(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s

The energy in ev/s

\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
3 years ago
Other questions:
  • A weight lifter is trying to lift a 1500-N weight but can apply a force of only 1200 N on the weight. One of his friends helps h
    10·1 answer
  • A loop antenna of area A = 3.04 cm^2 and resistance R = 6.66 μΩ is perpendicular to a uniform magnetic field of magnitude 18.4 μ
    13·1 answer
  • A suitable model for air could be b-b's moving rapidly around and colliding with each other.
    14·2 answers
  • Page 423 in the 8th grade science fusion textbook what at the answers
    10·1 answer
  • If a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?
    6·1 answer
  • An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
    11·1 answer
  • To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to
    5·1 answer
  • A 66.0-kg boy and his 45.0-kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending
    11·1 answer
  • Please help will mark brainliest!
    15·2 answers
  • What was the first and largest asteroid to be identified?.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!