Answer 2 or 3 questions.?

6.
(a)
Electrical energy, E
= P t
= 60 × (8 × 3600) J
= 1730000 J
= 1730 kJ

(b)
Electrical energy converted to light
= 1730 × 15% kJ
= 2600 kJ

(c)
It is converted to heat.

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7.
(a)
Efficiency
= (70/120) × 100%
= 58%

(b)
K.E. of the baseball in J:
(1/2) m v² = 70
(1/2) (0.145) v² = 70
Speed of the baseball, v = 31.1 m/s

====
(a)
Potential energy is converted to kinetic energy.

(b)
Take g = 9.8 m/s²
Efficiency
= [(K.E. at the bottom of the hill)/(P.E. at the top of the hill)] × 100%
= {[(1/2) m v²]/(m g h)} × 100%
= [v²/(2 g h)] × 100%
= [25²/(2 × 9.8 × 75)] × 100%
= 42.5%

(c)
It is used as work to overcome the frictional force.

6
a) E = 0.06*8 = 0.48 kwh
b) E' = 0.48*0.15 = 0.072 kwh
c) Heat is the proper answer

7
a) efficiency ɳ = 100*70/120 = 700/12 = 58.3 %
b) V = √ 2E/m = √ 140/0.145 = 31 m/sec 

8
a) from GPE to KE and Heat due to friction 
b)
GPE = m*g*H = 65*9,807*75 = 47800 joule
KE = m/2*V^2 = 32.5*25^2 = 20310 joule
eff.cy ɳ = 100*20310 / 47800 = 42.5%
c) It becomes heat

6 (a) Electrical energy = 60W * 8.0hr = 480 W-hr
(b) Radiant energy = 15% x Electrical energy = 0.15*480 W-hr = 72 W-hr
(c) Converts into thermal energy

7 (a) Efficiency = (70J/120J) = 0.5833 = 58.33%
(b) KE = ½mv²
v² = 2KE/m
v = √(2KE/m)
v = √(2*70/0.145)
v = 31 m/s

8 (a) Potential energy coverts to Kinetic energy
(b) PE = (65 kg)(9.81 m/s²)(75m) = 47823.75 J
KE = ½mv² = ½(65 kg) (25 m/s)² = 20312.5 J

Efficiency = (PE - KE)/PE
= (20312.5 J)/(47823.75 J)
= 0.42.47
= 42.47%

(c) It was used up by friction

What does 60 W mean? ( 60 joules every second)  How many seconds are there?  That gives you the energy used.  And if only 15% of that goes into light then what is 15% of your previous answer? I don't know when you studied percentages but I think I did them in grades 5 and 6.  You need to go back to those grade 6 maths notebooks and revise that work.