Answer:
W ≈ 5.9 × J
Hence, option (c) is correct.
Given:
mass of box = 6 kg
Displacement of box = 10 m
To find:
Work done by the force = ?
Formula used:
Work done is equal to change is potential energy,
W = m g h
Where, W = Work done
m = mass of the box
h = displacement by the force
Solution:
Work done is equal to change is potential energy,
W = m g h
Where, W = Work done
m = mass of the box
h = displacement by the force
W = 6 × 9.8 ×10
W = 588 Joule
Thus, W ≈ 5.9 × J
Hence, option (c) is correct.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A =
where d is the diameter of the wire
1.613 x 10^-7 =
6.448 x 10^-7 = 3.142 x
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>