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Hatshy [7]
3 years ago
7

Why are the 4 properties (cohesive behavior, ability moderate temperatures, expansion upon freezing, + versatility as a solvent)

of water important to life? ​
Physics
1 answer:
mylen [45]3 years ago
8 0

Answer:

Water is essential to life because of four important properties: cohesion and adhesion, water's high specific heat, water's ability to expand when frozen, and its ability to dissolve a wide variety of substances.

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The state of matter that has particles in a fixed position, with a definte volume and shape is
Rainbow [258]
That's the description of the SOLID phase of matter.
7 0
3 years ago
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±
gladu [14]

Answer:

E(final)/E(initial)=2

Explanation:

Applying the law of gauss to two parallel plates with  charge density equal σ:

E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2

3 0
3 years ago
Which statement is true about the force of gravity?
kumpel [21]

Answer:

The moon's gravity pulls the Earth to make tides.

Explanation:

The Moons Gravity Pulls On The Earth With Different  Strenght Making High Tide And Low Tide.

Hope This Helps!

4 0
2 years ago
The football player running towards the goal line has
blagie [28]

the football player has speed

8 0
3 years ago
Read 2 more answers
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
3 years ago
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