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Olegator [25]
3 years ago
8

Which event may occur when ocean salinity increases?

Physics
2 answers:
Travka [436]3 years ago
6 0

Answer:

sunlight in deeper location decreases

Explanation:

because the more salinity water has the harder it is to see and the harder it is for light to travel through the ocean

miv72 [106K]3 years ago
3 0
Sunlight in deeper locations decrease
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A triangle ∆P QR has vertices P(3, 2, −4), Q(1, 0, −4), R(2, 1, 1). Use the distance formula to decide which one of the followin
777dan777 [17]

Answer:

a. FALSE

b.TRUE

C. FALSE

Explanation:

The formula fot the distance between two points is given as

d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} +(z_{2}-z_{1})^{2}}\\

hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

PQ=\sqrt{(1-3)^{2} +(0-2)^{2} +(-4-(-4))^{2}}\\PQ=\sqrt{4+4+0}\\ PQ=\sqrt{8}

For point PR

we have

PR=\sqrt{(2-3)^{2} +(1-2)^{2} +(1-(-4))^{2}}\\PR=\sqrt{1+1+9}\\ PR=\sqrt{11}\\

|PQ|\neq |PR|

B. For point RP

RP=\sqrt{(3-2)^{2} +(2-1)^{2} +(-4-1)^{2}}\\RP=\sqrt{1+1+25}\\ RP=\sqrt{27}

for point RQ  we have

RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}

|RP|=|R Q|

C.

QP=\sqrt{(3-1)^{2} +(2-0)^{2} +(-4+4)^{2}}\\QP=\sqrt{4+4+0}\\ QP=\sqrt{8}

For point Q R

QR=\sqrt{(2-1)^{2} +(1-0)^{2} +(1-(-4))^{2}}\\QR=\sqrt{1+1+9}\\ QR=\sqrt{11}\\

QP\neq QR

6 0
3 years ago
A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at rrr = 4.0 mm is πr
max2010maxim [7]

Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is  \phi _2  = 1.57 \  rad

Explanation:

From the question we are told that

    The wavelength of the spherical wave is  \lambda =  2.0 \ mm =  \frac{2}{1000} = 0.002 \ m

    The first radius  is  r_1  = 4.0 \ mm  = \frac{4}{1000}  = 0.004 \ m

     The phase at that instant is  \phi _1 = \pi \ rad

     The second radius is  r_2  = 3.5 \ mm  = \frac{3.5}{1000}  = 0.0035 \ m

Generally the phase difference is mathematically represented as

          \Delta  \phi =  \phi _2 -  \phi _1

this can also be expressed as

         \Delta \phi =  \frac{2 \pi }{\lambda } (r_2 - r_1 )

So we have that

   \phi _2 -  \phi _1 =   \frac{2 \pi }{\lambda } (r_2 - r_1 )

substituting values

     \phi _2 -  \pi =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )

    \phi _2  =   \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) +   3.142

   \phi _2  = 1.57 \  rad

7 0
3 years ago
At a certain location, the horizontal component of the earth's magnetic field is 2.5 × 10−5 T, due north. A proton moves eastwar
Bogdan [553]

Answer:

  v = 4.10 10⁻³ m / s

Explanation:

For this exercise we will use Newton's second law where the force is magnetic

       F -W = m a

As the field is directed to the north and the proton to the east, using the rule of the right hand the force is vertical upwards, the force balances the weight the acceleration is zero

      F = W

      q v B = m g

Let's calculate the speed

      v = m g / (q B)

      v = 1,673 10⁻²⁷ 9.8 / (1.6 10⁻¹⁹ 2.5 10⁻⁵)

       v = 4.10 10⁻³ m / s

7 0
4 years ago
a 1385 kg cannon containing a 58.5 kg cannon ball is on wheels. the cannon fires the cannon ball giving it a velocity of 49.8 m/
lys-0071 [83]
Conservation of momentum requires that the sum of momenta after is equal to that before.  Since initially nothing is moving, the sum after the shot will also add to zero. 
m₁v₁ = -m₂v₂

Solve for the cannon's velocity v₁

v₁ = -m₂v₂/m₁ = -2.10m/s

The negative sign means it's moving 2.10m/s south.
7 0
3 years ago
What conclusion can be drawn from the statement that an element has high electron affinity, high electronegativity, and a high i
amid [387]

Answer:

D) The element is most likely from Group 6A or 7A and in period 2 or 3.

Explanation:

Electronegativity of an atom is the tendency of an atom to attract shared paired of electron to itself. Electronegativity increase across the period from left to right.The ability of an atom to attract electron to itself is electronegativity. Group 7A and 6A elements can easily attract atoms to itself so they are highly electronegative. The most electronegative element in the periodic table is fluorine.Group 6A and 7A is likely to have high electronegativity.

Electron affinity of an atom is the amount of energy release when an atom gains electron . Generally, when atom gains electron they become negatively charged. Group 6A and 7A elements have high electron affinity.  

Ionization energy is the energy required to remove one or more electron from a neutral atom to form cations.  ionization energy of group 7A and 6A are usually high because the energy required to remove these electron is usually very high . The elements in this groups usually gain electron easily so the energy to remove electron is very high.

4 0
3 years ago
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