A 3.00 kg box is lifted by an upward force 1.50 m above the surface pf Earth to the top of a table. What is the potential energy
1 answer:
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Answer:
Energy needed = 54.02 J
Explanation:
the Energy in an elastic spring from hookes law is given as
F= ke , therefore the energy (E) is
E =
K = 19.5 N/cm
e = 1.39cm
E = x 19.5 x 1.39
E = 13.55 J
The energy to stretch the spring for 6.93cm is
E = x 19.5 x 6.93
E = 67.57 J
The more energy needed for the further stretch is
67.57 - 13.55
Energy needed = 54.02 J
Answer:
Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T
Explanation:
Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :
......(1)
Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.
In this problem,
Current, I = 0.7 A
Length of wire, L = 0.62 m
Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m
Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m
Substitute these values in equation (1).
B = 6.99 x 10⁻⁶ T
Answer:
The radius = 170.21 m
Explanation:
The given data are : -
The centripetal acceleration of a car = 9.40 m/s².
Speed of a car = 40.0 m/s .
We have to calculate the radius ( r ) of of curve.
The centripetal acceleration ( a ) is given by
a =
r = =
=
= 170.21 m