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3 years ago

A 20-gauge shell is placed in a 12-gauge shotgun. What is the danger?

Physics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

Explanation:

If 20 gauge shell is placed in a 12 gauge shotgun it will cause damage and serious injury. If 20 gauge shell is placed the it will slip pass the chamber and and when fired it will cause ersonal damage. So to avoid the damage you will have to see what gauge is written on the shell and check if it is same as on gun barrel. Be careful in choice of ammunition.

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An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

BlackZzzverrR [31]

Answer:

a) $s = 0.603\,m$ , b) $s = 2.412\,m$

Explanation:

a) The system ice cube-spring is modelled by means of the Principle of Energy Conservation. Let assume that height at the bottom is zero:

$U_{g} = U_{k}$

$m\cdot g \cdot s\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x^{2}$

$s = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \sin \theta}$

$s = \frac{(25\,\frac{N}{m} )\cdot (0.1\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 0.603\,m$

b) The distance travelled by the ice cube is:

$s = \frac{(25\,\frac{N}{m} )\cdot (0.2\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 2.412\,m$

7 0

3 years ago

Read 2 more answers

In this lab, you will use a magnet and a simple circuit to examine the concepts of electricity and magnetism. Think about how th

Alex73 [517]

Sample Response: How can magnetic and electric fields be demonstrated?

5 0

3 years ago

Read 2 more answers

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0

2 years ago

As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change

Umnica [9.8K]

Refer to the diagram shown below.

m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

6 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

3 years ago