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:
no:
Explanation:
it would grow and no longer be able to fit through the loop due to the hot air expanding.
Answer: The angle between force and displacement should be θ = 90° for minimum work. The angle between force and displacement should be θ = 0° for maximum work.
Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:

substitute the values to find the speed of train:

Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s