The frictional force on the body is 0.25 N the net force on the body is 5.95 N while the acceleration produced is 2.83 m/s^2
<h3>What is the net force?</h3>
The net force is used to describe the resultant force that acts on the object.
The frictional force that acts on the object s obtained from;
Ff = 0.15 * 2.1-kg * 9.8 m/s^2
Ff = 0.25 N
The net force that acts on the body = F - Ff where F is the applied the force
= 6.2 N - 0.25 N
Net force = 5.95 N
Now;
Net force = ma
a= Net force /m
a = 5.95 N/2.1-kg
a = 2.83 m/s^2
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It will be then a positive ion and will be written as,
Al^3+
It’s hay, most farmers have their livestock eat hay and grass
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Yo sup??
Average velocity=total distance covered/total time taken
total distance covered=4 + 8=12 miles
total time taken=6 hours
Therefore
average velocity=12/6
=2 miles/hour
Hope this helps
