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azamat
3 years ago
5

WHO WANts TO HAVE SOME GLIZZY ACTION

Physics
1 answer:
Lilit [14]3 years ago
6 0
Bruh huh.............
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Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.
arlik [135]

Answer:

Unusually large moons form in giant impacts, which are relatively rare events

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible. The transition from an undifferentiated cloud to a star system complete with planets and moons takes about 100 million years.

8 0
3 years ago
An open bed truck is moving at 20 m/s and comes to a stop. A 100 kg crate sitting in the back of the truck remains at rest in th
zzz [600]

The work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

<h3>Work done by friction to prevent the crate from moving</h3>

The work done is determined by applying the principle of conservation of energy.

W = K.E

W = ¹/₂mv²

where;

  • m is mass of crate
  • v is speed of the crate with respect to the truck = 20 m/s

W = ¹/₂(100)(20²)

W = 20,000 J

Thus, the work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

3 0
2 years ago
Morse code and string phone cup
polet [3.4K]

Answer:

Following are the responses to the given question:

Explanation:

Following are the difference in the speed and accuracy:

The biggest difference was its tone between both the cup method as well as the Morse code. One should talk using cup system and sequence system and let its covering be understood in Morse code, just a series of beeps weren’t tangible... Secondly, a method of cup involves a range string. Rukun Negara is often distributed by cable, but this is a greater range. It reaches concurrently in those other areas.

In the case of using the string method for the block, its distance and stiffness of the string are limited and the barriers are eliminated. It will certainly be convenient and clearer if one utilizes Sign language.

If two experts are acting like such a machine on both sides, their information between both the 2 persons is smoother and quicker as they will not have to wait for data from a third person which may create another pause at the time.

Morse code is faster, eventually. For some very small periods when sensitive information is not transmitted, strings or cup methods must be reserved.

8 0
3 years ago
A block is launched up a frictionless 40° slope with an initial speed v and reaches a maximum vertical height h. The same block
Strike441 [17]

Answer: 1. h

Explanation:

The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.

Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.

Try to use the formula;

1/2MV2 = mgh

Where V = √(2gh)

I hope this helps

3 0
3 years ago
Hydraulic systems utilize Pascal's principle by transmitting pressure from one cylinder (called the primary) to another (called
Svetradugi [14.3K]

Answer:

The force to be applied on the primary piston is 140.63 N.

Explanation:

To solve this problem we apply the following formulas:

Pascal principle: F=P*A   Formula (1)

F=Force applied to the piston

P: Pressure

A= Piston area

A=\frac{\pi*d^{2} }{4} Formula (2)

d= piston diameter

Nomenclature:

Fp= Force on the primary piston

Fs= Force on the secondary piston

Ap= Primary piston area

As= Secondary piston area

W= car weight

Calculation of the force Fp necessary to support the weight of the car.

Pascal principle: Fp=P*Ap (Equation1)

In (Equation1) we know Ap and we don't know P.

Pressure calculation:

We apply Newton's first law for a balanced Secondary piston -automobile system

Newton's first law: ∑F=0

W-Fs=0

W=Fs

W=P*As

P=\frac{W}{As}

We replace P=\frac{W}{As}in equation 1:

Fp=\frac{W}{As} *Ap

Fp=\frac{W}{\frac{\pi*d_{s}^{2}   }{4} } *\pi *\frac{d_{p^{2} } }{4}

Fp=W*\frac{d_{p}^{2}  }{d_{s}^{2}  }

Fp=2200*\frac{2,1^{2} }{26^{2} }

Fp=2200*\frac{4.41}{676}

Fp=14.35kg

Calculation of Fp in Newtons (N):

1kg=9.8N

Fp=14.35kg*\frac{9.8N}{kg}

Fp=140.63N

3 0
3 years ago
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