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Hatshy [7]
1 year ago
11

A film badge worn by a radiologist indicates that she has received an absorbed dose of 2. 5 x 10-6 gy. the mass of the radiologi

st is 71 kg. how much energy has she absorbed?
Physics
1 answer:
Anestetic [448]1 year ago
4 0

Energy absorbed by the radiologist is 1.775*10^-6J.

To find the answer, we have to know about the radiation.

<h3>How much energy has she absorbed?</h3>
  • We have the expression for dose of absorption as,

           D=\frac{E}{m}

where; E is the energy absorbed and m is the mass of the body.

  • From the above expression, substituting appropriate values given in the question, we get,

         E=D*m=2.5*10^{-6}J/kg*71kg=1.775*10^{-4}J

Thus, we can conclude that, the Energy absorbed by the radiologist is 1.775*10^-6J.

<h3 />

Learn more about the radiation here:

brainly.com/question/24491547

#SPJ4

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A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin
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(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

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S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

v_c t=S_0 + v_g t

(v_c -v_g t)=S_0

t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: S_c = v_c t =(27.2 m/s)(7.5 s)=204 m

Gazelle: S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m

So, the gazelle should be ahead of the cheetah of at least

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A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant
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Answer:

9.96\cdot 10^{-10}J

Explanation:

The capacitance of the parallel-plate capacitor is given by

C=\epsilon_0 k \frac{A}{d}

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2 is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F

Now we can calculate the energy of the capacitor, given by:

U=\frac{1}{2}CV^2

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Substituting,

U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

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