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GuDViN [60]
3 years ago
14

Mr. Tolman believes that our universe is expanding, but with all of the gravitational force from the celestial bodies of space,

the universe will begin to contract. He believes in the _____.
A) oscillating universe theory
B) inflation theory
C) steady state theory
D) big bang theory
Physics
1 answer:
sattari [20]3 years ago
8 0
The correct answer is A I believe but I may not be correct. Either A or B
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Answer: The power is a measure of the rate at which work is done (or similarly, at which energy is transferred). The standard unit of power is Watt (1W). The power is determined by the change in energy, delta E (number of joules) and delta t, which is the time taken in seconds then: P= delta E/ delta t.

Mark me as Brainliest

Explanation:

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Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita
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Answer:

6.49 x 10^-8 N

Explanation:

formula is

F= G * ((m1 * m2)/r^2)

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3 years ago
Explain the energy transformations that occur when breaking in a hybrid<br> vehicle.
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3 years ago
Read 2 more answers
4-12. The morning inspection of the tank farm finds a leak in the turpentine tank. The leak is repaired. An investigation finds
bonufazy [111]

Answer:

a) V=759.8727\ ft^3

b) \dot V=1.403\times 10^{-3}\ ft^3.s^{-1}

c) t\approx29.541\ s

Explanation:

Given:

  • diameter of hole in the tank, d'=0.1\ in=\frac{1}{120}\ ft
  • position of the hole form the tank bottom, h' =7\ ft
  • initial level of turpentine in the tank before the leakage, h_i=17.3\ ft
  • level of turpentine in the tank after the repair of leakage, h_f=13\ ft
  • diameter of the tank, d=15\ ft
  • density of turpentine oil, \rho=55\ lbm.ft^3

a)

Now, volume of turpentine spilled:

V=A.(h_i-h_f)

where:

A= area of the cross section of the tank's volume

V=\pi .\frac{d^2}{4} \times(h_i-h_f)

V=\pi\times\frac{15^2}{4} \times(17.3-13)

V=759.8727\ ft^3

b)

When the tank was full the liquid level was highest:

so velocity form the height of the hole will be given as:

v=\sqrt{2g.(h_i-h')}

v=\sqrt{2\times 32.12\times (17.3-7) }

v=25.722\ ft.s^{-1}

<u>Now we have the flow rate of the spillage given by:</u>

\dot V=(\pi.\frac{d'^2}{4}) \times v

\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722

\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}

c)

Total time the leak was active can be calculated as:

t=\frac{V}{\dot V}

t=\frac{759.8727}{25.722}

t\approx29.541\ s

7 0
4 years ago
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