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Alex17521 [72]
3 years ago
13

If a 76.2 kg patient is exposed to 52.6 rad of radiation from a beta source, then what is the dose (mrem) absorbed by the person

's body?
Physics
1 answer:
Paha777 [63]3 years ago
7 0

Answer:

The dose absorbed by the person's body is 52600 mrem.

Explanation:

Given that,

Radiation = 52.6 rad

We need to calculate the absorbed dose

We know that,

The equivalent dose is equal to the absorbed radiation for beta source.

So, The patient is exposed 52.6 rad of radiation.

Therefore, The absorbed radiation is equal to the exposed 52.6 rad of radiation.

1 rad = \dfrac{1\ rem}{1000\ mrem}

So, radiation absorbed = 52.6 rad

radiation\ absorbed =52.6\times1000\ mrem

radiation\ absorbed = 52600\ mrem

Hence, The dose absorbed by the person's body is 52600 mrem.

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A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the
Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

m_1v = m_1v_1 + m_2v_2

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute m_2 = m_1/2 \& v_1 = v/3

m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}

Divide both side by m_1, then multiply by 6 we have

6v = 2v + 3v_2

3v_2 = 4v

v_2 = \frac{4v}{3}

So the final speed of the second car is 4/3 of the first car original speed

5 0
3 years ago
An aircraft acceleration from 100m/s to 300m/s in 100 s what is acceleration​
Crank

Answer:

(300  - 100) \div 100

8 0
3 years ago
The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the
Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

  • the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

average \ speed = \frac{total \ distance }{total \ time }

The total distance from the graph is calculated as follows;

  • first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
  • first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
  • second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
  • second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
  • third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
  • fourth downward distance from 12 cm to 9 cm = 3 cm
  • final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

  • This shortest distance is the straight line connecting the start and end position. Call this line P
  • From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
  • Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

                                                ↓

  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm

The average velocity of the ant is calculated as;

average \ velocity= \frac{\Delta displacemnt  }{total\ time }= \frac{14.318 \ cm}{105 \  s} = 0.136 \ cm/s  \\\\

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0
2 years ago
1. 1500j of work was done to move a box 20m. What force was applied to the box ?
Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

2. Potential Energy, PE = mass x gravity x change in height

we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

4 0
3 years ago
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