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Andrej [43]
3 years ago
13

You stop for a cappuccino at a coffee shop and notice that the tiny white bubbles of steamed milk remain on the surface of the c

offee. These air-filled bubbles stay where they are, rather than descending into the coffee or rising into the air, because they are:
A. more dense than the coffee but less dense than the air above thecoffee.

B. thicker than the coffee but less thick than the air above thecoffee.

C. less dense than the coffee but more dense than the air above thecoffee.

D. lighter than the cup of coffee but heavier than the column ofair above the coffee.
Physics
1 answer:
Rom4ik [11]3 years ago
3 0

Answer:

C. less dense than the coffee but more dense than the air above thecoffee.

Explanation:

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What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!
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Explanation:

Multiply 657 by 32:

 657

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⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

   ↳            21024 g/mL

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In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o
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3 0
3 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

  • The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s  

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

The final linear momentum will be:

\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

so:

 \vec{P}_0= \vec{P}_f 

( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

so

\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

So, for the truck

m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

And, for the car

950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

5 0
3 years ago
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