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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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A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>

Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

$Ro = m/V$

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

$V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]$

And the mass m = 4 [gramm] = 0.004 [kg]

$Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]$

3 0

3 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

3 years ago

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

3 years ago

I need help on this.

4vir4ik [10]

It brings cold water from the bottom of the ocean.

4 0

3 years ago

Read 2 more answers

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0

3 years ago