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deff fn [24]
3 years ago
9

What two factors determine the density of water in deep currents?

Physics
1 answer:
Alecsey [184]3 years ago
7 0

Answer:

There are two main factors that makeocean water more or less dense than about 1027 kg/m3: the temperature of the water and the salinity of the water.Ocean water gets more dense as temperature goes down. So, the colder the water, the more dense it is. Increasing salinity also increases thedensity of sea water.

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Which kind of energy does the soccer player transfer to the ball?
djyliett [7]

1. Kinetic

He makes the ball move by kicking it, which increases the kinetic energy

4 0
3 years ago
The fuzzy cloud around the nucleus is called the
Kamila [148]
It's number three on your worksheet. ;)
6 0
3 years ago
How to not get heart attacks ? And what is some plans/list for a healthy diet ?
Mamont248 [21]

Answer:

hey there

( i )we have to Choose good nutrition. A healthy diet is one of the best weapons you have to fight cardiovascular disease. and should be physically active and stop smoking if you do these are some

life style changes though which can prevent heart attacks

( ii )An healthy diet for heart

lots of fruits and vegetables.

nuts, beans, and legumes.

whole grains.

plant-based oils, such as olive oil.

low-fat dairy products.

Explanation:

Hope it helps you

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6 0
3 years ago
How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?​
miss Akunina [59]

Answer:

The answer to your question is    Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

           Pe = mgh

Substitution

           Pe = (50)(5)(9.81)

Simplification

           Pe = 2452.5 J

8 0
3 years ago
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th
Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0
3 years ago
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