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3 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Physics

2 answers:

sergey [27]3 years ago

7 0

Answer:

1)The mass of water increases.

2) Freezing point of water decreases.

4) bouyancy of objects in the water increases.

Salinity in the oceanic water is attributed by the water coming from other sources like rivers carrying rocks and soil containing salt. Salinity decreases the temperature of water, making it cold and more denser. The cold water settles at the bottom of the ocean. It flows across the oceanic basins as slow currents. The objects being less in density flows over the surface of water. Hence,bouyancy of objects in water increases.

Pavlova-9 [17]3 years ago

3 0

The answer is number two, number four, and number one

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if the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspe

Mazyrski [523]

If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspeed will Increase and true altitude will increase.

<h3>What is Altitude?</h3>

This refers to the vertical elevation of a body and is the measurement of the height above the sea or ground level. An airplane's altitude is directly proportional to the airspeed as long as the power is constant.

This means that increase in airspeed will lead to an increase in altitude and vie versa. This is the reason why as it runs fast, it can be observed that it begins to elevate.

Read more about Altitude here brainly.com/question/1159693

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1 year ago

A 60 newton force is directed in a southwest direction. What are the magnitudes of the force components in the south and west di

JulijaS [17]

Hello
This question is to be tackled using vectors. When we look at the Southwest direction, we know it lies exactly between the South and West directions, producing an angle of 45 degrees. When we take sin(45) and multiply it by the original force, we obtain the component towards West; that is:
60 * sin (45) = 42.4 Newtons
And cos(45) gives us the force towards the South direction; that is:
60 * cos(45) = 42.4 Newtons
This can also be checked by using the formula of the magnitude of a vector and squaring 42.4, adding it to the square of 42.4 and then taking the square root of the answer.
sqrt(42.4^2 + 42.4^2) = sqrt(3600) = 60

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3 years ago

A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us

Rina8888 [55]

Answer:

The magnetic induction of the magnetic field is 0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒ the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can consider the current density as the total current over the whole area:

I(enclosed) = I / (πr ²) * πr' ²

I(enclosed) = (I* r'²)/ (r ²)

with I = total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire 0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0 * (I* r'²)/ (r ²)

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant = 4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') = μ 0 * (I* r'²)/ (r ²)

Because the circumference of a circle is 2 π r , when we integrate over length at a distance r ′ from the center of wire whose crossection is a circle we get 2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B = 4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T = 0.0005293 mT

The magnetic induction of the magnetic field is 0.0005293 mT

6 0

3 years ago

In a living organism, a fixed fraction 1.30 × 10-12 of 12C is the radioactive isotope 14C, which has a half life of 5730 y.

laila [671]

Answer:

A = 0.325 Bq

Explanation:

given,

half life of ¹⁴C = 5730 years

fixed fraction 1.30 × 10⁻¹² of ¹²C

half life = 5730 years

T_{1/2} = 5730 x 365 x 24 x 60 x 60

= 1.807 x 10¹¹ s

radioactive decay constant

λ = $\dfrac{0.693}{T_{1/2}}$

λ = $\dfrac{0.693}{1.807\times 10^{11}}$

λ = 3.835 x 10⁻¹² /s

number of atom

mass m = 1.30 g

$n = \dfrac{m_0}{M}\times 6.022 \times 10^{23}$

$n = \dfrac{1.30}{12}\times 6.022 \times 10^{23}$

n = 6.524 x 10²²

Number of ¹⁴C atoms in 1.3 g of sample N ' =

= 1.30 × 10⁻¹² x 6.524 x 10²²

= 8.482 x 10¹⁰

Required activity, A = λ N

A = 3.835 x 10⁻¹² x 8.482 x 10¹⁰

A = 0.325 Bq

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erastova [34]

Answer:

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3 years ago