Gravity attracts an object to

Guys please helpp!!!!1

Setler79 [48]

Answer:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy ( $E$ ) is the sum of gravitational potential energy ( $U$ ) and translational kinetic energy ( $K$ ), all measured in joules. In addition, gravitational potential energy is directly proportional to height ( $h$ ) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

$K = E$ , $U = 0$

Position B/Position D

$K = (1-x)\cdot E$ , $U = x\cdot E$ , for $0 < x < 1$

Position C

$K = 0$ , $U = E$

3 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

If you were thinking about a washing machine as a system which of the following represents the inputs?

Lostsunrise [7]

Answer:

The answer is a, the dirty cloths, water and detergent.

Explanation:

The answer is the above selected because the inputs basically represent the data that are passed through the system to generate the output.

In this case, the inputs are the aforementioned in the answer while the possible output would literally be the clean cloths.

4 0

3 years ago

Please help ASAP A uniform disk of mass M and radius R is pivoted such that it can rotate freely about a horizontal axis through

kompoz [17]

Answer: F = mg(1 + 4m / (½M + m))

Explanation:

"At this point seems" unclear. If the particle is at the top of the disc and angular velocity is negligible, then the force would equal the weight of the particle. F = mg

The more interesting question would be what force is needed to keep the particle attached when significant angular rotation has been achieved. The maximum point would be diametrically opposed to the starting point.

I will analyze it there

The potential energy will convert to kinetic energy

mgh = ½Iω²

mg(2R) = ½(½MR² + mR²)ω²

4mgR = R²(½M + m)ω²

ω² = 4mg / (R(½M + m))

With m at the lowest position, the force of attachment must support the weight of m and provide for the needed centripetal acceleration

F = m(g + ω²R)

F = m(g + 4mg / (R(½M + m))R)

F = mg(1 + 4m / (½M + m))

7 0

2 years ago

What is the equation for elastic potential energy?

drek231 [11]

Answer: HOPE THIS HELPED YOU! :D ;P

Elastic potential energy = force x distance of displacement.

Explanation:

6 0

3 years ago

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