There is more wire to travel through,farther distance, and a higher possibility of other disruptions. Please Mark Brainliest!!!
Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method
Explanation:
Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:
--> Surface EMT and
--> Intramuscular EMT
Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.
Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.
The difference between the two forms of EMT includes:
- surface EMT is non- invasive while intramuscular EMT is invasive
- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.
- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.
Answer:
Yes, but only if it's sunny.
Explanation:
As you know, solar panels generate energy through the sun's rays of light (better known as sunlight). Therefore, as long as the sun is shining high in the sky, the car will generate electricity and be able to function. If this vehicle was only powered by solar panels, it would not function during the night, in cloudy areas, and/or in dark places (such as parking garages or home garages).
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The image of the water tower and the houses is in the attachment.
Answer: (a) P = 245kPa;
(b) P = 173.5 kPa
Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.
The pressure is calculated as: P = hρg
where
ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;
When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:
ρ = 
V = ρ/m
V = 
V = 525 m³
To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:
V = 
![r = \sqrt[3]{ \frac{3}{4\pi } .V}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%7D%7B4%5Cpi%20%7D%20.V%7D)
r = ![\sqrt[3]{\frac{525.3}{4\pi } }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Cfrac%7B525.3%7D%7B4%5Cpi%20%7D%20%7D)
r = 5.005 m
diameter = 2*r = 10.01m
(a) Height for House A:
h = 15 + 10.01
h = 25.01
P = hρg
P = 25.01.10³.9.8
P = 245.10³ Pa or 245kPa
(b) h = 25 - 7.3
h = 17.71
P = hρg
P = 17.71.1000.9.8
P = 173.5.10³ Pa or 173.5 kPa