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Ilya [14]
3 years ago
14

If the Lifted Index is positive, is the parcel warmer than, the same temperature as, or cooler than its environment?

Physics
1 answer:
OLga [1]3 years ago
8 0

Answer:

A positive lifted index means that a parcel of air, if lifted, would be COLDER than the surrounding air at 500mb. The air is, therefore, stable and would resist vertical motion. Large positive values (+8) would normally indicate very stable air.

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Your answer is option "D. feline"

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Choose all of the items that are incorrectly matched. A) inactivation gate — closed at restB) activation gate — open at restC) i
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What energy transformations take place in Joule's system?
attashe74 [19]

The gravitational potential energy of the weight changes to motion energy when the weight moves down and the paddle wheel spins. The motion energy changes to thermal energy, causing the water’s temperature to rise.

(fast explanation for my fellow edmentum kids :))

6 0
4 years ago
Read 2 more answers
A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling
sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

P_{1} = P_{2}

m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

v_{1_{i}}: is the initial velocity of the ball 1 = 6.20 m/s

v_{2_{i}}: is the initial velocity of the ball 2 = 0 (it is at rest)

v_{1_{f}}: is the final velocity of the ball 1 =?

v_{2_{f}}: is the initial velocity of the ball 2 =?

m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}

v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} (1)        

Now, by conservation of kinetic energy (since they collide elastically):

\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}          

m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}  (2)

By entering equation (1) into (2) we have:

m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}    

0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}            

By solving the above equation for v_{2_{f}}:

v_{2_{f}} = 3.1 m/s

Now, v_{1_{f}} can be calculated with equation (1):

v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s

The minus sign of v_{1_{f}} means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!                  

5 0
3 years ago
Henry is trying to move his 232 kg refrigerator. He is pushing, but the refrigerator is not moving. If the maximum coefficient o
Inessa [10]

Answer:

Once the fridge is moving you are still applying the same push force but now the opposing friction (now kinetic) force is less.

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