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IRINA_888 [86]
2 years ago
10

Air escaping out from an air hose at a gas station always feels cool. Why? ​

Physics
1 answer:
kati45 [8]2 years ago
8 0

Answer:

The air is contained at a high pressure in the tube. When it escapes from a small orifice, it suddenly expands. A large amount of its heat is absorbed in the process of expansion resulting in considerable fall in its temperature. This is why the escaping air feels cold.

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The measure of how much salt will dissolve into 100g of water is _______________ .
Bingel [31]
The measure of how much salt will dissolve into 100g of water is _solution_ .
8 0
3 years ago
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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
3 years ago
Which one of newton's laws does a doll riding a dog represent
strojnjashka [21]
I believe it would best represent Newton’s first law; an object tends to stay at rest and an object tends to stay in motion unless acted upon by an unbalanced force. When the dog stops walking, the doll will continue to go forward because there is no unbalanced force acting in it.
4 0
3 years ago
At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at
love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

brainly.com/question/18557256

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brainly.com/question/17088562

4 0
3 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
3 years ago
Read 2 more answers
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