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kondor19780726 [428]

3 years ago

13

As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the

floor become warm. During this process, what happens to energy?

2 answers:

Galina-37 [17]3 years ago

7 0

Hello! The answer would be conserved.
I hope this helped:)
Mark me brainilest if you get the chance please!
Have a great

san4es73 [151]3 years ago

3 0

Answer:

conserved

Explanation:

During this process the energy is conserved

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

Objects floating in the water, like buoys, only bob up and down when waves pass. Why do they not get pushed all the way to where

weeeeeb [17]

Answer:

Because as the waves propagates, the particles of the medium (molecules of water) vibrates perpendicularly (upward and downward) about their mean position and not in the direction of the waves.

Explanation:

A wave is a phenomena which causes a disturbance in a medium without any permanent deformation to the medium. Examples are; transverse wave and longitudinal wave. Waves transfer energy from one point in the medium to another.

The waves generated by water are transverse waves. Which are waves in which the vibrations of the particles of the medium is perpendicular to the direction of propagation of the waves.

Thus as the waves propagates, the molecules of water vibrates up and down and not along the direction of propagation of the waves. So that the floating objects do not get pushed in the direction of the waves every time.

8 0

3 years ago

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would

natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a) $v_{es} =v_{rms}$ = 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given, $v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

7 0

3 years ago

What is the volume of the cone?

dezoksy [38]

Answer:

42.417 cm³

Explanation:

The formula to find the volume of a cone is :

V = $\frac{1}{3}$ × π r² h

Here,

r ⇒ radius ⇒ 3 cm

h ⇒ height ⇒ 4.5 cm

<u>Let us find it now.</u>

V = $\frac{1}{3}$ × π r² h

V = $\frac{1}{3}$ × π × 3 × 3 × 4.5

V = $\frac{1}{3}$ × π × 9 × 4.5

V = $\frac{1}{3}$ × π × 9 × 4.5

V = $\frac{1}{3}$ × π × 40.5

V = $\frac{1}{3}$ × 3.142 × 40.5

V = $\frac{1}{3}$ × 127.251

V = <u>42.417 cm³</u>

4 0

2 years ago

3x/y=6g/b solve for x

lora16 [44]

So looking at the problem, you are going to want to start by finding a common denominator (1) in this case: yb, and combining like terms (2). You are then going to want to multiply both sides by (yb) as the reciprocal to the fractions (3).
1) 3x 6g
---- = ---
y b

2) 3xb 6gy
------ = -----
yb yb

3) 3xb 6gy
(yb) ------ = -----
yb yb
which becomes: 3xb = 6gy

So after this, things become much more simple, as all you have to do is isolate the (x), which can be done by dividing the entire equation by (3b).

3xb 6gy
----- = -----
3b 3b

where you will then find your answer of:
2gy
x = ----- (simplified by the GCM of 3)
b

5 0

3 years ago