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salantis [7]
3 years ago
10

During a medical evaluation, the doctor can __________.

Engineering
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

Treat the patient

i hope this is ur answer

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D
TEA [102]

Answer:

true

Explanation:

4 0
3 years ago
An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
s344n2d4d5 [400]

Answer:

Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.  

Explanation:

3 0
3 years ago
What is the braks mean effictive pressure?
OverLord2011 [107]

Engine cylinder pressure

<u>Explanation:</u>

  1. Brake mean effective pressure is a method to calculate  the engine cylinder pressure which  would give the measured brake horsepower. Brake mean effective pressure is used to identify engine efficiency regardless of capacity or engine speed.
  2. It is used to identify engine efficiency.It is measured by means of transducers or pressure gauges.
  3. It is the measure of engine capacity to do work.

7 0
3 years ago
The Chinese had a toy helicopter some 1000 years ago. True Or False ?
Vadim26 [7]

Answer:

f

Explanation:

for the lazy people

5 0
3 years ago
A microwave transmitter has an output of 0.1W at 2 GHz. Assume that this transmitter is used in a microwave communication system
Len [333]

Answer:

gain = 353.3616

P_r = 1.742*10^-8 W

Explanation:

Given:

- The output Power P_o = 0.1 W

- The diameter of the antennas d = 1.2 m

- The frequency of signal f = 2 GHz

Find:

a. What is the gain of each antenna?

b. If the receiving antenna is located 24 km from the transmitting antenna over a free space path, find the available signal power out of the receiving antenna.

Solution:

- The gain of the parabolic antenna is given by the following formula:

                            gain = 0.56 * 4 * pi^2 * r^2 / λ^2

Where, λ : The wavelength of signal

            r: Radius of antenna = d / 2 = 1.2 / 2 = 0.6 m

- The wavelength can be determined by:

                            λ = c / f

                            λ = (3*10^8) / (2*10^9)

                            λ = 0.15 m

- Plug in the values in the gain formula:

                            gain = 0.56 * 4 * pi^2 * 0.6^2 / 0.15^2

                            gain = 353.3616

- The available signal power out from the receiving antenna is:

                            P_r = (gain^2 * λ^2 * W) / (16*pi^2 * 10^2 * 10^6)

                            P_r = (353.36^2 * 0.15^2 * 0.1) / (16*pi^2 * 10^2 * 10^6)

                            P_r = 1.742*10^-8 W

4 0
3 years ago
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