Answer:
490.5 N
Explanation:
Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction
F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.
F=kmg=0.5*100*9.81=490.5 N
Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
Answer:
Tissues that are damaged or injured.
Explanation:
Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.
Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.
Hey there!
The answer to your question is: Iron
There is a very high concentration of iron and other metals in the mesosphere. There is a further buildup of these metals when meteorites burn up in the atmosphere and release such metal atoms.
Thank you!
Potential difference required in an electron microscope to give an electron wavelength of 4. 5 nm will be 0.063 V.
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other is called potential difference.
The wavelength of an electron is calculated for a given energy (accelerating voltage) by using the de Broglie relation between the momentum p and the wavelength λ of an electron
lambda = 4.5 nm = 4.5 * 
m
h = 
J s
e = 1.6 * 
C
m = 9.1 * 
kg
Energy = eV
lambda = h / 
= h / 
= 
/ (2m (eV))
V = / (2 m e )
V = 
/ 2 * 9.1 * * 1.6 * * 
V = 0.063 V
To learn more about wavelength of an electron here
brainly.com/question/17295250
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