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Talja [164]
4 years ago
5

A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly a

llowable effective dose did she receive?
Physics
1 answer:
jonny [76]4 years ago
6 0

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad

Thus, the effective dose allowable yearly = 16 times

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An unstoppable object is heading right toward an unmovable object. What's going to happen?
marshall27 [118]

Answer:

In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.

<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
8 0
2 years ago
A power boat pulls a water skier 2.6 km maintaining a
Luba_88 [7]

Answer:

W = 650 [kJ]

Explanation:

The definition of work is denoted by the product of force by the distance traveled by the body, this distance traveled corresponds to the direction of the force.

In this case we have:

d = distance = 2.6[km] = 2600 [m]

F = force = 250 [N]

W = F*d = 250 * 2600 = 650000 [J] or 650 [kJ]

7 0
3 years ago
The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
snow_lady [41]
Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0
3 years ago
Read 2 more answers
Contrary to popular belief, a ski jumper does not achieve a large amount of "air" when doing a jump (less than 6 feet). This is
Elina [12.6K]

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, V_x = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

d_y = V_yt + \frac{1}{2}gt^2

initial vertical velocity, V_y, = 0

d_y =  \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

6 0
3 years ago
Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1
harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation E_p=h f, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of E_N=N h f.

We also know that frequency and wavelength are related by f=\frac{c}{\lambda}, so we have E_N=\frac{N h c}{\lambda}, where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

N=\frac{\lambda E_N}{h c}

We need to know then how much energy do we have to calculate N. The equation of power is P=E/t, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be E_N.

Putting this paragraph in equations:

E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J.

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}

3 0
3 years ago
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