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3 years ago

Why do X-Ray technicians stand behind a shield when giving X-Rays to patients?

2 answers:

7 0

Lol I just learned this, here's the short answer:
Large doses of X-Rays are known to cause cell damage and cancer, so
X-Ray technicians have to stand behind the shield. Here's the lengthy answer:
 This is because the patient would only be exposed to the small dose of radiation once, which is hardly harmful. However, radiographers deal with this all day every day, which could mean that the small doses could add up. This is why they stand behind a lead screen; the small exposure to the X-Rays aren't harmful alone but frequent exposure could begin to damage your cells and lead to potential genetic mutation. If a patient is often in radiography (a long-stay patient for example), they begin to wear lead aprons whilst being X-rayed in order to prevent the risk of harm. Hope I answered correctly.

Mashcka [7]3 years ago

7 0

Answer:

because large doses of X-Rays are known to cause cell damage and cancers

Explanation:^^

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Which is a way to improve self esteem?

erik [133]

Answer: Positive Talk

Explanation:

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3 years ago

Someone help me in science plz

NARA [144]

Answer:

I would say Climate - A

Explanation:

Just looks like the logical thing.

7 0

2 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

radius of the curve, r = 90 m
angle of inclination, θ = 10.8⁰
speed of the car, v = 75 km/h = 20.83 m/s
mass of the car, m = 1100 kg

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0

2 years ago

A woman on a bridge 95.6 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a

NARA [144]

Answer:

The speed of the raft is 1.05 m/s

Explanation:

The equation for the position of the stone is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

The equation for the position of the raft is as follows:

x = x0 + v · t

Where:

x = position of the raft at time t

x0 = initial position

v = velocity

t = time

To find the speed of the raft, we have to know how much time the raft traveled until the stone reached the river. For that, we can calculate the time of free fall of the stone:

y = y0 + v0 · t + 1/2 · g · t² (v0=0 because the stone is dropped from rest)

If we place the origin of the frame of reference at the river below the bridge:

0 m = 95.6 m - 9.8 m/s² · t²

-95.6 m / -9,8 m/s² = t²

t = 3.12 s

We know that the raft traveled (4.84 m - 1.56 m) 3.28 m in that time, then the velocity of the raft will be:

x/t = v

3.28 m / 3.12 s = v

v = 1.05 m/s

5 0

2 years ago

Read 2 more answers

A ball of mass 1.200 kg is attached to a 0.700 m long string. It is moving horizontally in a clockwise direction, 4.000 m above

ankoles [38]

Answer:

The ball is making a circular motion. The centripetal acceleration in a circular motion is

$a = \frac{v^2}{r}$

$a = \frac{20^2}{0.7} = \frac{400}{0.7} = 571.4~ m/s^2$ ~

Explanation:

Normally in physics questions like this, we use all the information in the question. Here, we left out the mass or the height of the object. I assume there is more sub-questions that is not shared here. However, the centripetal acceleration can be found by using the velocity and the radius only.

7 0

3 years ago