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rusak2 [61]
3 years ago
14

At room​ temperature, glass used in windows actually has some properties of a liquid. it has a very​ slow, viscous flow.​ (visco

sity is the property of a fluid that resists internal flow. for​ example, lemonade flows more easily than fudge syrup. fudge syrup has a higher viscosity than​ lemonade.) glass does not become a true liquid until temperatures are greater than or equal to 515515​°c. find the fahrenheit temperatures for which glass is a liquid. ​hint: use the formula
Physics
1 answer:
tamaranim1 [39]3 years ago
8 0
<span>The fahrenheit temperature is 927965. It is calculated using the formula 515515 Degree Cx1.8+32=927965. The degree celcius and fahrenheit are two units two measure temperature. If the value is given in celcius it can be converted into fahrenheit using the above formula.</span>
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How does radiation differ from conduction?
Natali5045456 [20]

Answer:

b

Explanation:

3 0
3 years ago
Un avión de rescate en Alaska deja caer un paquete de provisiones a un grupo de exploradores extraviados. Si el avión viaja hori
posledela

Answer:

180.4 m

Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

$V_{ox}=v_x = 40 \ m/s$

   h = 100 m  

    x =?

     Height formula h:

     $h=g \times \frac{t^2}{2}$

      Time t is cleared:

     $t = \sqrt{\frac{2h}{g}}$

      $t = \sqrt{\frac{2 \times 100}{9.8}}$

      t = 4.51 sec

 Horizontal distance formula x:

       $x=V_x \times t$

        x = 40 m / sec x 4.51 sec

        x = 180.4 m

4 0
3 years ago
A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete
ale4655 [162]

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0
3 years ago
What is the current in a series circuit that has two resistors (4.0 ohms and
Maslowich
C 1.6 amps hipe this helps
6 0
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.

7 0
3 years ago
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