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3 years ago

6

The water cycle involves water in the liquid, solid, and gaseous states of matter. In which of the following Earth systems can w

ater be found in one or more of these states?

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

in the hydrosphere

Explanation:

solid=glaciers

liquid=ocean

vapor=clouds

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The potential energy between two atoms in a particular molecule has the form U(x) = 2.1 x 8 − 5.2 x4 where the units of x are le

a_sh-v [17]

Answer:

$x\approx 0.948$

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

$U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}$

Where $x$ is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

$F = -\frac{dU}{dx}$

The function is derived in terms of distance:

$F (x) = \frac{84}{5\cdot x^{9}} -\frac{104}{5\cdot x^{5}}$

Then, it is needed to find at least of x so that F(x) equals to 0.

$\frac{84}{5\cdot x^{9}}-\frac{104}{5\cdot x^{5}}=0$

$\frac{84}{x^{4}}-104 = 0$

$84-104\cdot x^{4} = 0$

$x=\sqrt[4]{\frac{84}{104} }$

$x\approx 0.948$

7 0

3 years ago

9 m/s to a total stop. what is the objects acceleration

MAVERICK [17]

We would need to know the time it took to slow to a stop.

5 0

3 years ago

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.1 m/s. The coefficient of kinetic fri

Lelu [443]

Answer:

114.19186 m

Explanation:

v = Velocity of girl at bottom = 7.1 m/s

$\mu$ = Coefficient of kinetic friction = 0.045

g = Acceleration due to gravity = 9.81 m/s²

d = Distance

N = Normal force = 783 N

The weight of the sled and girl is considered as the sum of the frictional force and weight

Hence we use the following equation where the kinetic energy and potential energy are conserved

$\frac{1}{2}mv^2=\mu Nd\\\Rightarrow \frac{1}{2}\frac{783}{9.81}\times 7.1^2=0.045\times 783\times d\\\Rightarrow d=\frac{1}{2}\times \frac{783\times 7.1^2}{9.81\times 0.045\times 783}\\\Rightarrow d=114.19186\ m$

The sled travels 114.19186 m on the level ground before coming to a rest

5 0

3 years ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

A cardiac defibrillator stores 1275 J of energy when it is charged to 5.6kV What is the capacitance? O 11.3 pF 96.1 pF O 813 F

Angelina_Jolie [31]

Answer: $813.13(10)^{-7}F$

Explanation:

The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor $W$ and its capacitance $C$ , which is given by the following equation:

$W=\frac{1}{2}CV^{2}$ (1)

Where:

$W=1275J$

$V=5.6kV=5.6(10)^{3}V$ is the voltage

$C$ is the capacitance in Farads, the value we want to find

Isolating $C$ from (1):

$C=\frac{2W}{V^{2}}$ (2)

$C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}$ (3)

Finally:

$C=0.000081313F=813.13(10)^{-7}F$ This is the capacitance of the cardiac defibrillator

5 0

3 years ago