Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
Answer:
Explanation:
Given data
Mass m=1.2 g=0.0012 kg
diameter d=0.76 m
Friction Force F=3.6 N
To find
Velocity v
Solution
From the Centripetal force we know that
Where m is mass
v is velocity
r is radius
Substitute the given values to find velocity v
So
Answer:
As per Coulomb's law we know that force between two charges is given as
F = \frac{kq_1q_2}{r^2}F=
r
2
kq
1
q
2
here we know that
q_1 = 2.5 \times 10^{-6} Cq
1
=2.5×10
−6
C
q_2 = -5.0 \times 10^{-6} Cq
2
=−5.0×10
−6
C
r = 0.0050 mr=0.0050m
now from above formula we will have
F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}F=
(0.0050)
2
(9×10
9
)(2.5×10
−6
)(5×10
−6
)
F = 4500 NF=4500N
so they will attract towards each other as they are opposite in nature with force F = 4500 N
Answer:
Mecanical into electrical,kinectic and thermal energy
