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bearhunter [10]

3 years ago

9

If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0 ∘C to 85.0 ∘C ?

1 answer:

andrey2020 [161]3 years ago

3 0

Answer:

Explanation:

$C_{water}$ = 4190 J/kg.K

$C_{Al}$ = 910 J/Kg. K

$m_{Al}$ = 1.50 kg

$m_{water}$ = 1.80 kg

$Q_{added} = Q_{Al} + Q_{water}$

$=m_{Al} C_{Al}$ ΔT + $m_{water} C_{water}$ ΔT

= (1.50)(910)(85.0-20)+(1.80)(4190)(85.0-20)

= 578,955 J

= 579 kJ

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Answer:

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Explanation:

From the question we are told that

The mass of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

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The height attained by the combined mass is $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation

$KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as

$PE_m = [m_b + m_w] * g * h$

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

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Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

so

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

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