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3 years ago

What might cause a shotgun to explode? this is due in like 10 mins so help me please

1 answer:

5 0

The terminating pin strikes the preliminary, making it detonate. The flash from the groundwork touches off the black powder. Gas changed over from the consuming powder quickly extends in the cartridge. ... The shot's speed and getting away from gases produce a "blast."

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The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand

dsp73

P1v1/t1 = p2v2/t2
p1=475, v1=4, t1=290
v2=6.5, t2=277
solve for p2 in kpa

5 0

3 years ago

Convert the following dB to decimal a. -12 b. 3 c. 10 d. 0

stiks02 [169]

Answer:

for -12db

$\frac{V1}{V2}=0.251\\\\\frac{P1}{P2}=0.0631$

for 3db

$\frac{V1}{V2}=\sqrt{2}\\\\\frac{P1}{P2}=2$

for 10db

$\frac{V1}{V2}=3.16\\\\\frac{P1}{P2}=10$

for 0db

$\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1$

Explanation:

The decibel is a logaritmic value given by:

$db=10*log(\frac{P1}{P2})=20*log(\frac{V1}{V2})$

we use 10 for power values and 20 for other values such voltages or currents.

$\frac{V1}{V2}=10^{\frac{db}{10}}\\\\\frac{P1}{P2}=10^{\frac{db}{20}}$

for -12db

$\frac{V1}{V2}=10^{\frac{-12}{10}}=0.251\\\\\frac{P1}{P2}=10^{\frac{-12}{20}}=0.0631$

for 3db

$\frac{V1}{V2}=10^{\frac{3}{10}}=\sqrt{2}\\\\\frac{P1}{P2}=10^{\frac{3}{20}}=2$

for 10db

$\frac{V1}{V2}=10^{\frac{10}{10}}=3.16\\\\\frac{P1}{P2}=10^{\frac{10}{20}}=10$

for 0db

$\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1$

4 0

3 years ago

If the magnitude of the initial velocity of the ball v0 = 7.94 ± 0.03, and the "gun" is tilted 31 ± 0.4º upwards, what is the un

ycow [4]

Answer:

0.05

Explanation:

Given a variable C which is the product of two variables A, B:

$C=A\cdot B$

Then the absolute error on C is given by:

$\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}$

where $\sigma_A, \sigma_B, \sigma_C$ are the uncertanties on the measure of A, B and C, respectively.

In this problem, the horizontal component of the velocity $v_x$ is given by

$v_x = v_0 cos \theta$

Therefore, the uncertainty on vx is given by:

$\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta}$ (1)

where we have:

$v_0 = 7.94$

$\sigma_{v_0}=0.03$

$\theta=31^{\circ}$ , so

$cos \theta=cos 31^{\circ}=0.857$

The uncertainty on $cos \theta$ is given by:

$\sigma_{cos \theta}=|sin \theta|\sigma_\theta$

where:

$|sin \theta| = |sin 31^{\circ}|=0.515$

and

$\sigma_\theta=0.4^{\circ}=0.007 rad$

$\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036$

Also,

$v_x = v_0 cos \theta = (7.94)(cos 31^{\circ})=6.81 m/s$

So, combininb everything into (1), we find:

$\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05$

4 0

3 years ago

HELPPPPPP PLEASE thanks

Artist 52 [7]

Your answer is letter A

7 0

3 years ago

Read 2 more answers