Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N -
=
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Answer:
i beleave cuz of the Earth is spherical
Explanation:
Explanation:
Unclear question. The clear rendering reads;
"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?
Answer:
a)= 98kJ
b)=108kJ
c) = 10kJ
Explanation:
a. The work that is done by gravity on the elevator is:
Work = force * distance
= mass * gravity * distance
= 1000 * 9.81 * 10
= 98,000 J
= 98kJ
b)The net force equation in the cable
T - mg = ma
T = m(g+a)
T = 1000(9.8 + 10)
T = 10800N
The work done by the cable is
W = T × d
= 10800N × 10
= 108000
=108kJ
c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J
Work done by cable = PE +KE
108,100 J = KE + 98,100 J
KE = 10,000 J
= 10kJ
=
Answer:
Diameter will be 351.42 mm
Explanation:
We have given current flowing in the copper wire i = 310 A
Voltage drop across the wire V = 0.55 volt
We know that resistance is given by 
Length of the copper wire l = 1 m
Resisitivity of the copper wire 
We know that resistance 


As area 


So diameter
= 351.42 mm