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3241004551 [841]
3 years ago
12

If you weighed 100 lb on Earth, what would you weigh at the upper atmosphere of Jupiter? For reference, Jupiter has a mass that

is about 300 times Earth’s mass and a radius that is 10 times Earth’s radius.
Physics
1 answer:
nignag [31]3 years ago
7 0

Answer:

The answer to the question is

A 100 lb person would weigh 300.33 lbf at the upper atmosphere of Jupiter

Explanation:

To solve the question we note that

Mass of object = 100 lb =‪ 45.35924‬ kg

Mass of Jupiter = 300×Mass of Earth = 300×5.972 × 10²⁴ kg =1.7916×10²⁷ kg

Radius of Jupiter = 10× Radius of Earth = 10×6,371 km = 63710 km

Gravitational constant, G = 6.67408 × 10⁻¹¹ m³ kg-1 s-2

Gravitational force is given by F_G= \frac{Gm_1m_2}{r^2}

Plugging in the values we get

F_G = \frac{6.67408*10^{-11}*45.35924*1.7916*10^{27}}{63710^2} = 1335.93 N

Converting into lbf gives 1335.93 N *0.2248 lbf/N = 300.33 lbf

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3 years ago
a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

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3 years ago
An atom emits a photon when one of its electrons:
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Answer:

undergoes a transition to a quantum state of lower energy

Explanation:

When electrons in an atom move to another quantum state, they emit/absorb a photon according to the following:

- If the electron is moving to a higher energy state, it absorbs a photon (because it needs energy to move to a higher energy level, so it must absorb the energy of the photon)

- if the electron is moving to a lower energy state, it emits a photon (because it releases the excess energy)

In particular, the energy of the absorbed/emitted photon is exactly equal to the difference in energy between the two levels of the electron transition:

E=|E_1 - E_2|

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What is Electromagnet?<br><br> What is Voltage?
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Answer: a core of magnetic material

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8 0
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If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same
Angelina_Jolie [31]

Answer:W=\frac{3}{4} ft-lb

Explanation:

Given

Work required to stretch 1 ft is 12 ft-lb

and we have to find work required to stretch 3 in.

i.e. \frac{1}{4} ft

12=\frac{1}{2}K\left ( 1\right )^2 ------(1)

W=\frac{1}{2}k\left ( \frac{1}{4}\right )^2-----(2)

divide (1)&(2)

\frac{12}{W}=\left ( \frac{4}{1}\right )^2

W=\frac{12}{4\times 4}

W=\frac{3}{4} ft-lb

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