Given :
Mass of box , m = 250 kg.
Force applied , F = 285 N.
The value of the incline angle is 30°.
the coefficient of dynamic friction is .
To Find :
The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.
Solution :
Net force applied in box is :
Acceleration , .
By equation of motion :
Therefore, the speed of box is 12.04 m/s.
Hence, this is the required solution.
Answer:
The resistivity of the wire is
Explanation:
We have,
Length of a wire is 40 cm or 0.4 m
Diameter of a wire is 0.60 mm
Radius is 0.3 mm or 0.0003 m
Resistance of a wire is 1.5 ohm
Now we need to find the resistivity of the material of which it is made. The resistance of a wire in terms of its resistance, length and area is given by :
= resistivity
So, the resistivity of the wire is .
Answer:
a) F = (137.4 i ^ + 185 j ^) N
b) F = 230.2 N , θ = 53.5º
Explanation:
In this exercise we ask to find the net force, for which we will define a coordinate system fix the donkey and use trigonometry to decompose the forces
Jack F₁ₓ = 63.9 N
Jill F₂ = 79.1 N with direction 45º to the left
cos (180 -45) = F₂ₓ / F₂
sin 135 = / F₂
F₂ₓ = F₂ cos 135
F_{2y} = F₂ sin 135
F₂ₓ = 79.1 cos 135 = -55.9 N
F_{2y} = 79.1 sin 135 = 55.9 N
Jane F₃ = 183 N direction 45th to the right
cos 45 = F₃ₓ / F3
sin 45 = F_{3y} / F3
F₃ₓ = F₃ cos 45 = 183 cos 45
F_{₃y} = F₃ sin 45 = 183 sin 45
F₃ₓ = 129.4 N
F_{3y} = 129.4 N
we add each component of the force
Fₓ = F₁ₓ + F₂ₓ + F₃ₓ
Fₓ = 63.9 + (-55.9) + 129.4
Fₓ = 137.4 N
F_{y} = F_{2y} + F_{3y}
F_{2y} = 55.9 + 129.4
F_{2y} = 185.3 N
we can give the result of the forms
a) F = (137.4 i ^ + 185 j ^) N
b) in the form of module and angle
F = RA (Fₓ² + F_{y}²)
F = Ra (137² + 185²)
F = 230.2 N
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (185/137)
θ = 53.5º