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erica [24]
3 years ago
6

Kalisha is going to cut the remaining 21 1/2 buns into thirds. How many 1/3 size buns will she have?

Physics
2 answers:
baherus [9]3 years ago
6 0
Total buns = 21 1/2 = 43/2 buns
1/3 size buns would be: 43/2 / 1/3 = 43*3 / 2 = 129/2

In short, Your Answer would be 129/2 or 64 1/2 or 64.5

Hope this helps!
solmaris [256]3 years ago
4 0
1/2 of 21 buns is 10.5. 1/3 of that is 3.5

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Please help! ASAP. I'm having a breakdown.
Marianna [84]

Answer:

What do you need help with?

4 0
3 years ago
(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
castortr0y [4]

Answer:

Explanation:

The word "sheet" implies that the copper is quite thin.

Copper is also a very good conductor of heat.

Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.

3 0
2 years ago
A police siren has a frequency of 340.0 Hz when the car is stationary and the temperature is 20°C. The police car at point S mov
hodyreva [135]

Answer:

a) 361.7 Hz

b) 318.1 Hz

c) 29.3 m/s

Explanation:

a) Speed of sound in 20 degrees celcius is 343 m/s.

Use v=wavelength*frequency to find wavelength.

343=wavelength*340

wavelegth=1.009 meters

Relative speed of waves to observer A = 343+22 = 365 m/s

Plug it back in.

365=1.009*F

F=361.7 Hz

b) Relative speed of waves to observer B = 343-22 = 321 m/s

Plug it back in.

321=1.009*F

F=318.1 Hz

c) STATIONARY CAR, so 369Hz=v/1.009. v=372.3, 372.3-343=29.3m/s!

8 0
3 years ago
Which planet do most known extrasolar planets most resemble?
tatuchka [14]
The awnser is A>Most known exoplanets resemble gas giants known as "hot Jupiters" as a result of being large objects orbiting close to they're host star.
Most known exoplanets that we have detected are likely or confirmed to be gas giants. This is because one of the most used detection techniques is the Transit technique.

This technique involves looking at a distance star's light curve (the changes in the brightness of star over a period of time). Most stars dim and brighten over time by a small bit. But if the light curve shows a periodic and large dip in brightness, this is a sign that something large is passing (or transiting) in front of it.

The reason most detected exoplanets are gas giants is simply because they're the easiest to detect. They cause a larger dip in the curve than a smaller planet would. A small planet's dip are masked by the star's normal dimming and brightening.

Imagine having a flashlight shining on a wall. If you pass a large object through the light beam, it has a large shadow. If you pass a smaller object through, it has a smaller shadow. The change in the light you see on the wall is similar to what you will see in a star's light curve. Now if you pass an object closer to flashlight, as opposed to closer to the wall, this changes the shadow as well.

This is why most exoplanets are "hot Jupiters". They're large gas giants that are hot because they pass very close to the star, resulting in a larger dip that is easier to see in the curve. They also have shorter orbits so they're more likely to be seen in a few weeks or month's time.


6 0
4 years ago
Figure 3 shows a bicycle of mass 15 kg resting in a vertical position, with the front and back
Vinil7 [7]

Explanation:

There are three forces on the bicycle:

Reaction force Rp pushing up at P,

Reaction force Rq pushing up at Q,

Weight force mg pulling down at O.

There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.

Sum of the forces in the y direction:

Rp + Rq − (15)(9.8) = 0

Rp + Rq − 147 = 0

Sum of the moments at P:

(15)(9.8)(0.30) − Rq(1) = 0

44.1 − Rq = 0

Sum of the moments at Q:

Rp(1) − (15)(9.8)(0.70) = 0

Rp − 102.9 = 0

Sum of the moments at O:

Rp(0.30) − Rq(0.70) = 0

0.3 Rp − 0.7 Rq = 0

Any combination of these equations will work.

3 0
3 years ago
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