All categories

3 years ago

What factors affect the chemical properties of seawater

1 answer:

3 0

An ocean's salinity is affected by water and air temperature, the volume of water that flows into it, its age and the chemical composition of surrounding geological formations.<span> Because the effect of these factors varies over the area of an ocean, different levels are found within the ocean</span>

You might be interested in

PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

horsena [70]

It will be cloudy and there will be rain.

Hope this helps

5 0

3 years ago

Read 2 more answers

A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato

Sergeu [11.5K]

Answer:

A) The speed of the potato at the lowest point of its motion is 7.004 m/s

B) The tension on the string at this point is 8.5347 N

Explanation:

Here we have that the height from which the potato is allowed to swing is 2.5 m

Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)

Where:

ω₂ = Final angular velocity

ω₁ = Initial angular velocity = 0 rad/s

α = Angular acceleration

θ₂ = Final angle position

θ₁ = Initial angle position

However, we have potential energy of the potato

= Mass m×Gravity g× Height h

= 0.29×9.81×2.5 = 7.1125 J

At he bottom of the swing, the potential energy will convert to kinetic energy as follows

K.E. = P.E. = 7.1125 J

1/2·m·v² = 7.1125 J

Therefore,

v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05

∴ v = √49.05 = 7.004 m/s

B) Here we have the tension given by

Tension T in the string = weight of potato + Radial force of motion

Weight of potato = mass of potato × gravity

Radial force of motion of potato = mass of potato × α,

where α = Angular acceleration = v²/r and r = length of the string

∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)

T = 8.5347 N

4 0

3 years ago

Read 2 more answers

What is your opinion on abortion?

zloy xaker [14]

Ratio

L nobody cares and bozo

5 0

3 years ago

Read 2 more answers

Can someone help me?

Setler [38]

Answer: (1, 30), (2,10), (3,40), (4,20)

Explanation:

4 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago