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Hatshy [7]
4 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]4 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.

You might be interested in
Definition of distance in physics
solniwko [45]

Answer:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Explanation:

8 0
3 years ago
WhAt is this question??????????
valina [46]

Answer:

Period = 7

Group = Actinides group

Family = Actinides Family

Explanation:

uranium found in seventh row or seventh period of the periodic table.

uranium is the member of Actinoides group it is also called Actinide group, And it has element whose atomic is number is 89 to 103. The atomic number of the uranium is 92. So, the uranium element is belongs to Actinide group

Uranium found in actinide family of periodic table, and actinides family has element whose atomic number is greater than or equal to 89 and less than or equal to 103.

8 0
3 years ago
A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is do
Alborosie

1. B. T

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}} (1)

where

L is the length of the pendulum

g is the gravitational acceleration

From the formula, we notice that the period of the pendulum does not depend on the mass of the bob. Therefore, when the bob's mass is doubled, the period does not change.

2. C: sqrt(6)*T

In this case, the pendulum is brought to the moon, where the gravitational acceleration is

g'=\frac{g}{6}

If we substitute this value into the equation for the period (1), we find the new period of the pendulum:

T'=2\pi \sqrt{\frac{L}{g'}}=2\pi \sqrt{\frac{L}{(g/6)}}=\sqrt{6}(2\pi \sqrt{\frac{L}{g}})=\sqrt{6} T

3. B: It will no longer oscillate because there is no gravity in space

Explanation:

The motion (oscillation) of the pendulum is caused by the force of gravity, which "pulls" the bob towards the equilibrium position. If there is no gravity, then there is no force acting on the bob, therefore the pendulum can no longer oscillate.

So, the correct answer is

B: It will no longer oscillate because there is no gravity in space

4 0
4 years ago
Read 2 more answers
Mary travelled 70mph due north
tiny-mole [99]
Velocity because It is defined as the change in the position with respect to the time. Velocity is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of velocity is, m/s

Acceleration : It is defined as the rate of change of velocity of an object wit respect to the time. Acceleration is a vector quantity that means it depends on the magnitude and direction of an object. The S.I unit of acceleration is, m/s to the power of 2

Distance : It is defined as the how far an object has traveled in time. Distance is a scalar quantity that means it is depends on the magnitude of an object only. The S.I unit of distance is, m

Speed : It is defined as the distance traveled by an object in unit time. Speed is also a scalar quantity. The S.I unit of speed is, m/s

Mary traveled 70 miles/hour due north. This is an example of velocity. 70 miles/hour tell us about the magnitude of the object and north tell us about the direction of an object.
Hence, the correct option is, velocity.







3 0
3 years ago
A typical person's eye is 2.5 cm in diameter and has a near point (the closest an object can be and still be seen in focus) of 2
Paraphin [41]

Answer:

2.27 cm

2.5 cm

Explanation:

u = Object distance =  25 cm

v = Image distance = 2.5 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{25}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{11}{25}\\\Rightarrow f=\frac{25}{11}=2.27\ cm

The minimum effective focal length of the focusing mechanism of the typical eye is 2.27 cm

when u=\infty

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{\infty}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{1}{2.5}\\\Rightarrow f=\frac{2.5}{1}=2.5\ cm

The maximum effective focal length of the focusing mechanism of the typical eye is 2.5 cm

3 0
3 years ago
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