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Llana [10]
3 years ago
13

Explain two ways in which water’s properties help sustain life. Earth Science

Physics
2 answers:
alexandr1967 [171]3 years ago
8 0

Answer:

capillarity and water’s relatively high specific heat are two properties of water that help sustain life. Capillarity is used by plants to get water from the ground through their roots to their leaves. Water’s high specific heat helps animals by helping regulate their body temperatures as they sweat or pant.

Explanation:

the sample answer on edge 2021

hjlf3 years ago
6 0

Answer:

Water is essential for all living things. Water's unique density, high specific heat, cohesion, adhesion, and solvent abilities allow it to support life.

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Find the resistance at 50°c of copper wire 2mm in diameter and 3m long
mr Goodwill [35]
0.0179 ohms for copper.

0.0184 ohms for annealed copper



Ď = R (A/l) where

Ď = electrical resistivity

R = electrical resistance of a uniform specimen

A = cross sectional area

l = length



Solve for R by multiplying both sides by l/A

R = Ď(l/A)



The cross section of the wire is pi * 1^2 mm = 3.14159 square mm = 3.14159e-6 square meters.

The length is 3 meters. So l/A = 3/3.14159e-6 = 9.5493e5



Ď for copper is 1.68e-8 so 1.68e-8 * 9.5493e5 = 1.60e-2 ohms at 20 C

But copper has a temperature coefficient (α) of 0.00386 per degree C.

So the resistance value needs to be adjusted based upon how far from 20 C the temperature is.

50 - 20 = 30 C

So 0.00386 * 30 = 0.1158 meaning that the actual resistance at 50 C will be 11.58% higher.

So 1.1158 * 0.016 = 0.0179 ohms.



If you're using annealed copper, the values for Ď and the temperature coefficient change.

Ď = 1.72e-8

α = 0.00393



Doing the math, you get

1.72e-8 * 9.5493e5 * (1 + 30 * 0.00393) = 0.0184 ohms
8 0
3 years ago
Help me please (*˘︶˘*).。*♡​
ss7ja [257]

Answer:

your answer is number 4

Explanation:

Hope it helps :)

pls mark brainliets :P

5 0
3 years ago
Read 2 more answers
A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6
Kitty [74]

Answer:

The velocity of the launcher after the projectile is launched is  -5.011 m/s

Explanation:

Here we have the mass of the cannon and cart, m₁ =  4.65 kg

Velocity of cannon and cart, v₁ = 2.00 m/s

Mass of projectile, m₂ = 50.0 g = 0.05 kg

Velocity of projectile, v₂ = 647 m/s

Velocity of the launcher, v₃ = Required

Mass of cannon and cart, launcher after launching projectile m₃ = 4.65-0.05

= 4.6 kg

Therefore, from the principle of the conservation of linear momentum, we have

Total initial momentum = Total final momentum

m₁ × v₁ = m₂ × v₂ + m₃ × v₃

Substituting gives

4.65 kg × 2.00 m/s = 0.05 kg × 647 m/s + 4.6 kg × v₃

4.65 kg × 2.00 m/s - 0.05 kg × 647 m/s = 4.6 kg × v₃

-23.05 kg·m/s = 4.6 kg × v₃

v_3 = \frac{-23.05 \, kg\cdot m/s}{4.6 \, kg} =  \frac{-461}{92} m/s

v₃ = -5.011 m/s.

3 0
3 years ago
Read 2 more answers
Facts about conolizing into mars?​
Leto [7]

Answer:

xxxx

Explanation:

Mars also has a thin atmosphere. Because of this it has potential to host humans and other organic life. That makes Mars the best choice for thriving colony off the earth. The moon has also been proposed as the first location for human colonization but it not known to have air or water.

Hope this helps!

Have an amazing day/night

7 0
3 years ago
On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L
vlabodo [156]

Answer:

75.6J

Explanation:

Hi!

To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,

Given the above we have the following equation.

Q=(m)(h2)-(m)(h1)

where

m=mass=1.3×10−3kg.

h2= entalpy at 37C

h1= entalpy at -20C

Q=m(h2-h1)

remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference

Q=mCp(T2-T1)

Cp= specific heat of air = 1020 J/kg⋅K

Q=(1.3×10−3)(1020)(37-(-20))=75.6J

4 0
3 years ago
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