Answer:
Explanation:
work done by applied force
= force x displacement
= 7.6 x 15 m
= 114 J .
Answer:
I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry
Explanation:
1.

x=5

7.
5,12,13
9.

Answer:
Part a) 
Part b) 
Explanation:
Part a) what is its frequency, in rev/s
we have that
An old-fashioned LP record rotates at 33 1/3 RPM
so

Convert mixed number to an improper fraction

Remember that

Convert rev/min to rev/sec

Simplify

Part b) what is it period, in seconds
we know that
The period is the reciprocal of the frequency
therefore
the frequency is

Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J