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2 years ago

7

Subsurface ocean currents continually circulate from the warm waters near the equator to the colder waters in other parts of the

world. What is the main cause of these currents?

1 answer:

soldi70 [24.7K]2 years ago

8 0

<span>the main cause of these currents is the </span>differences in density of ocean water.

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A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6

Shalnov [3]

Answer:

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

Explanation:

A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.

A) What is the wavefunction y(x,t) for the standing wave that is produced?

B) In which harmonic is the standing wave oscillating?

C) What is the frequency of the fundamental oscillation?

a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).

b. lambda=2L/n

when comparing the wave equation with the general wave equation , we get the wavelength to be

2*pi*x/lambda=6.98x

lambda=0.9m

we use the equation

lambda=2L/n

n=number of harmonics

L=length of string

0.9=2(1.35)/n

n=2.7/0.9

n=3

third harmonic

c. to calculate frequency , we compare with general wave equation

y(x,t)=Acos(kx+ωt)

from ωt=742t

ω=742

ω=2*pi*f

742/2*pi

f=118.09Hz

8 0

2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

B. Complete the table to show the effect of each change on each electric quantity. (1 point)

notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0

2 years ago

Select the correct answer from each drop-down menu.

bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons. An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

8 0

2 years ago

13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?

sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

$Q = mc del T$

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

$Q = 20*C*80$

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

5 0

2 years ago