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Fed [463]
4 years ago
7

A 49-year-old female was referred for mammography, but she is very apprehensive after reading about the risk of ionizing radiati

on. How should the radiographer handle this situation?
Physics
1 answer:
SOVA2 [1]4 years ago
4 0

Answer:

Mammography is the process in which low energy radiations are used to diagnose and screening. The purpose of this process is the early detection of the breast cancer. These low energy radiations may have some risks like damaging and burning of cells.  

In the current scenario, woman is apprehensive because she has read about the risks of using ionizing radiations. The radiographer should tell her the benefits of the mammography will outweigh its potential consequences. Screening, for instance, will let her know if she is suffering from breast cancer. Cancer is very dangerous disease as compare to very small burning.

In this way radiographer should handle the situation.

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Gravitational acceleration for the bodies falling freely is​
STatiana [176]

Answer:

g= 9.81m/s..................

3 0
3 years ago
A 1-kg plastic block is neutrally buoyant when submerged in water with a density of 1 g/cm3. If the same block is placed, such t
goldenfox [79]

Answer:

-2 m/s

Explanation:

The computation of the acceleration of the block is as follows:

The volume of the block is

V = 1kg ÷ 1,000 kg/m^3 = 1 ÷ 1,000 m^3

Now at the time of submerging in the oil

Net force is

m_g - F_B = m \times (-a)\\\\m_g - v\times S_{oil} \times g = -ma\\\\a = (\frac{v}{m} \times S_{oil}  - 1) g\\\\= (\frac{1}{1000} \times 0.8 \times 1000 -1 ) \times 10\\\\

= -2 m/s

8 0
3 years ago
It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you
natka813 [3]
If 56.5kJ are needed to raise the temp by 90°C and if the heater is 60% efficient that means that:
60% X y = 56.5kJ
where y is the electrical energy in kJ that the heater will use.
y = 94.2kJ 


6 0
3 years ago
A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?
Marizza181 [45]

Answer:

1.04%

Explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

\eta=\dfrac{P_o}{P_i}\times 100

Where

Po and Pi are output and input powers.

P_o=\dfrac{5000\times 10^3}{15}\\\\=3.34\times 10^5\ W

So,

\eta=\dfrac{3500}{3.34\times10^{5}}\times100\\\\=1.04\%

So, the percentage efficiency of the drill is 1.04%.

4 0
3 years ago
A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4
Brrunno [24]

Answer:

true

Explanation:

Area = πr²

Differentiating in respect to r

we get,

dA/dr = 2πr

using chain rule,

\frac{dA}{dt} = \frac{dA}{dr} .\frac{dr}{dt}  = 2\pi \frac{dr}{dt}

since r = 4feet

so ,

2π(4) = 8π

5 0
3 years ago
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