Answer:
the change in thermal energy of the projectile is 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
Draw a right triangle so that its hypotenuse is 600 ft. The adjacent side is below the vertical, and it makes an angle of 75° with the hypotenuse.
Let h = height of the right triangle.
By definition,
sin75° = h/600
h = 600*sin75° = 579.555 = 580 ft (nearest ft)
Answer: 580 ft (nearest foot)
The answer to your question will be C. because they are very inexpensive and are readily available but they will eventually deplete because we use them faster than they can be produced we use what has been building up.
Answer:
a)143.8 decays/minute
b)0.41 decays/minute
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of C-14= 5670 years
t= time taken to decay
Ao= activity of a living sample
A= activity of the sample under study
a)
0.693/5670 = 2.303/1000 log(162.5/A)
1.22×10^-4 = 2.303×10^-3 log(162.5/A)
1.22×10^-4/2.303×10^-3 = log(162.5/A)
0.53 × 10^-1 = log(162.5/A)
5.3 × 10^-2 = log(162.5/A)
162.5/A = Antilog (5.3 × 10^-2 )
A= 162.5/1.13
A= 143.8 decays/minute
b)
0.693/5670 = 2.303/50000 log(162.5/A)
1.22×10^-4 = 4.61×10^-5 log(162.5/A)
1.22×10^-4/4.61×10^-5 = log(162.5/A)
0.26 × 10^1 = log(162.5/A)
2.6= log(162.5/A)
162.5/A = Antilog (2.6 )
A= 162.5/398.1
A= 0.41 decays/minute
