(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
<h3>
Minimum force to be applied </h3>
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
- μ(s) is coefficient of static friction
- m is mass of the block
- g is acceleration due to gravity
F = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
<h3>Acceleration of the block</h3>
F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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Answer:
Tuesday bc instead of running he/she was walking bc he/she might not have as much energy
Explanation:
in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s
Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js
now 1 Joule= 10⁷ ergs
so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)
h=6.626 x10⁻²⁷ erg s
Explanation:
The weight of the car is equal to, 
...........(1)
Where
m is the mass of car
g is the acceleration due to gravity
The normal or vertical component of the force is, 
or
.............(2)
The horizontal component of the force is, 
Taking ratio of equation (1) and (2) as :
or
Hence, this is the required solution.
