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zzz [600]
3 years ago
5

A spring-loaded pellet gun is designed to fi re 3.00-g projectiles horizontally at a speed of 45.0 m/s. (a) If the spring is com

pressed to its maximum design difference of 8.00 cm, what spring constant is required?
Physics
1 answer:
taurus [48]3 years ago
3 0

Answer:

The spring constant required = 949.2 N/m

Explanation:

The energy stored in the spring is converted to kinetic energy of the projectiles fired.

Energy stored in spring's = kx²/2

Kinetic energy of bullets = mv²/2

k = spring constant = ?

x = maximum compression of the spring = 8 cm = 0.08 m

m = mass of projectiles fired = 3.0 g = 0.003 kg

v = velocity of projectiles fired = 45.0 m/s

kx² = mv²

k (0.08²) = 0.003 (45²) = 6.075

k = 949.2 N/m

The spring constant required = 949.2 N/m

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A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
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Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

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