Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U = 
................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 = 
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U = 
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
Answer:
a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C
, c) U = 3.21 10⁻¹¹ J
Explanation:
a) The capacitance of a capacitor is
C = k e₀ A / d
Let's calculate
C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²
C = 4,012 10⁻¹⁴ F
b) let's look the charge
C = Q / ΔV
Q = C ΔV
Q = 4,012 10⁻¹⁴ 400
Q = 1.6 10⁻¹¹ C
c) The stored energy
U = ½ C ΔV²
U = ½ 4,012 10⁻¹⁴ 400²
U = 3.21 10⁻¹¹ J
Answer:
intinal speed should be 10
Explanation:
v = v0 + at
30 = v0 +10.2
then v0= 10
Answer:
The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Explanation:
Given that,
Mass of halfback = 98 kg
Speed of halfback= 4.2 m/s
Mass of corner back = 85 kg
Speed of corner back = 5.5 m/s
We need to calculate their mutual speed immediately after the touchdown-saving tackle
Using conservation of momentum
Where, 
= mass of halfback
=mass of corner back
= velocity of halfback
= velocity of corner back
Put the value into the formula
Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2
