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,
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,
Thus, we can conclude that, the Energy absorbed by the radiologist is 1.775*10^-6J.
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Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
Answer:
Explanation:
The water particles just flow through each other. They cannot be destroyed nor created.
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Answer:
0.0021576N
Explanation:
F=(k)(q1q2/r^2)
F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)
F=0.0021576N
