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3 years ago

What is the length of AR?

Physics

1 answer:

AlladinOne [14]3 years ago

6 0

Answer:Assault rifle?

Explanation:

20 m16 size and a 14.5 barrel and get a pump shot and get the blue version and go to tilted and one pump even tho forkknife is garbage.:)

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tester [92]

Great question the answer is -25x.

3 0

3 years ago

A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres

Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0

3 years ago

g (12 points) The time between incoming phone calls at a call center is a random variable with exponential density p(x) = 1 r e

rusak2 [61]

Answer:

$(1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0$

Explanation:

A function f(x) is a Probability Density Function if it satisfies the following conditions:

$(1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0$

Given the function:

$p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}$

(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in $[0,\infty).$

(2)

$\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1$

The function p(x) satisfies the conditions for a probability density function.

6 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$