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2 years ago

Which property best makes radio waves safe for diagnosing illnesses through magnetic resonance imaging?

Physics

2 answers:

77julia77 [94]2 years ago

8 0

Answer:

1) Long energy of radio waves

2) long wavelengths of radio waves

Explanation:

aleksandrvk [35]2 years ago

4 0

Answer:

It is used in MRI because it does not damage cells

Radio waves are used for space research because they have very long wavelengths

Explanation:

Many parts of the electromagnetic spectrum are applied in clinical diagnosis and treatment of illnesses. However, these highly ionizing radiation damage cells and its dosage must be carefully managed to avoid creating radiation related health problems for the patients.

Radio waves can be used in MRI without issues because the energy of the radiation is not sufficient to cause damage to cells but is sufficient to provide images for the sake of medical diagnosis.

Secondly, radio waves have long wavelength. This property is suitable for long range

communication. Hence it can be used in space research

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The S-strain Pneumococcus bacteria had a smooth surface, because ____________________.

drek231 [11]

The S strain Pneumococcus bacteria had a smooth surface because IT IS SURROUNDED BY A CARBOHYDRATE CAPSULE CALLED THE S STRAIN. The other form, the R strain has a rough surface and no capsule. It is only the S strain that exhibits virulence.

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3 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

salantis [7]

Answer:

130m

Explanation:

You just have to multiply velocity by the time traveled:

100m/s * 1.3s = 130m!

5 0

3 years ago

Read 2 more answers

Which is an example of an unstructured activity that promotes resistance training?

aleksandrvk [35]

One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A.

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself.

3 0

2 years ago

A car is accelerated from 40 m/s to 48 m/s in 6.5 seconds. What is the magnitude of the car’s

pantera1 [17]

Answer:

Explanation:

AVerage acceleration is the cjange in velocity with time

a = v-u/t

v is the final velocity = 48m/s

u is the initial velocity = 40m/s

t is the time = 6.5s

a = 48-40/6.5

a = 8/6.5

a = 1.23m/s²

Hence the magnitude of the car’s average acceleration during this period is 1.23m/s²

4 0

2 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago