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arsen [322]
3 years ago
13

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

et while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.
Physics
1 answer:
muminat3 years ago
3 0

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.  

KE = 1/2*m*v^2

      = 1/2*0.005*320^2

      = 256 Joules.  

Divide this by the distance over which this energy was received and you have the force that provided that energy.  

  = 256/0.820 = 312.195 Newtons  

Rounded off, this is 312 N

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3 years ago
In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

8 0
3 years ago
Read 2 more answers
A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means o
Delicious77 [7]

Answer:

b) N = 560 N, c)  fr = 138.56 N, d)  μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

              N + T_y -W = 0

              N = W -T_y

let's use trigonometry for tension

             sin θ = T_y / T

             cos θ = Tₓ / T

             T_y = T sin θ

             Tₓ = T cos θ

we substitute

              N = W - T sin 30

we calculate

              N = 640 - 160 sin 30

              N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

              Tₓ - fr = 0

               Tₓ = fr

we substitute and calculate

              fr = 160 cos 30

              fr = 138.56 N

d) the friction force has the formula

             fr = μ N

             μ = fr / N

we calculate

             μ = 138.56 / 560

             μ = 0.247

4 0
3 years ago
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