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Nuetrik [128]
3 years ago
10

Research is being done on the use of radio waves in destroying cancer cells. What type of frequency would be best used in this t

echnology?
Physics
1 answer:
shusha [124]3 years ago
8 0

Answer:

Radiofrequency Ablation (RFA) which is a type of Electrical Energy is used to treat cancer. If it sees heat made by radio waves to kill cancer cells. Ablation means destroying completely. The method is putting RFA a probe (electrode) that goes through the skin into the tumor. A CT scan is needed so to check that the probe is in exactly the right place. An electrode in the probe creates radiofrequency energy to produce heat and kill the cancer cells.

Explanation:

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Answer:

Here's an explanation but not the answer

Explanation:

When a resistor is traversed in the same direction as the current, the ... Traversing the internal resistance r1 from c to d gives −I2r1. ... I1 = I2 + I3 = (6−2I1) + (22.5− 3I1) = 28.5 − 5I1.

3 0
3 years ago
Arrange the steps of energy production in the correct order.
Ulleksa [173]
Here are the correct steps involved in energy production: 

<span>Step 1 - Consumption of food
Step 2 - breakdown of starch into glucose
Step 3 - absorption of glucose molecules
</span>Step 4 -cellular respiration in mitochondria<span>
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4 years ago
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Laundry in a clothes dryer often becomes charged with static electricity while drying. Which of these best explains why a clothe
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C- Electrons are transferred as clothes rub against each other in the dryer.
7 0
3 years ago
Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl
WINSTONCH [101]

In circular motion we know that there is two type of acceleration that Cheryl experience

1. Tangential acceleration

2. Centripetal acceleration

here given that

a_t = 3 m/s^2

for centripetal acceleration we know that

a_c = \frac{v^2}{R}

a_c = \frac{4^2}{2} = 8 m/s^2

now we know that both centripetal acceleration and tangential acceleration is perpendicular to each other

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a_{net} = \sqrt{8^2 + 3^2} = 8.54 m/s^2

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8 0
3 years ago
A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0
3 years ago
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