If the coefficient of static friction is 0.3, then the minimum force required to get it moving is equal in magnitude to the maximum static friction that can hold the body in place.
By Newton's second law,
• the net vertical force is 0, since the body doesn't move up or down, and in particular
∑ <em>F</em> = <em>n</em> - <em>mg</em> = <em>n</em> - 50 N = 0 ==> <em>n</em> = 50 N
where <em>n</em> is the magnitude of the normal force; and
• the net horizontal force is also 0, since static friction keeps the body from moving, with
∑ <em>F</em> = <em>F'</em> - <em>f</em> = <em>F'</em> - <em>µn</em> = <em>F'</em> - 0.3 (50 N) = 0 ==> <em>F'</em> = 15 N
where <em>F'</em> is the magnitude of the applied force, <em>f</em> is the magnitude of static friction, and <em>µ</em> is the friction coefficient.
Answer:
V = (5.8cm/s)i, (4.7cm/s)j
Explanation:
Given :
r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^
To obtain the average velocity (V)
V = (r2 - r1) / (t2 - t1)
To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above
r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j
r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j
r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j
r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j
r2 = (16.1cm)i + (9.4cm)j
V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0
V = 11.6i / 2 ; 9.4j / 2
V = (5.8cm/s)i, (4.7cm/s)j
The correct option is D.
The most effective solution to the scenario given in the question, is to appoint professionals who will locate new phosphate rich areas on the land. This will allow the residents of the area to use that particular area for their agricultural activities.
Answer:
The linear mass density of rope is 0.16 kg/m.
Explanation:
mass, m = 0.52 kg
force, F = 47 N
length, L = 3.3 m
(a) The linear mass density of the rope is defined as the mass of the rope per unit length.
Linear mass density = m/L = 0.52/3.3 = 0.16 kg/m
