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Maru [420]
4 years ago
15

The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty

in its momentum is 0.20%, how accurately can the position of the bullet be measured along the direction of motion?
Physics
1 answer:
nydimaria [60]4 years ago
7 0

Answer:

Uncertainty in position of the bullet is \Delta x=1.07\times 10^{-33}\ m

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

p=mv

p=0.035\times 709=24.81\ kg-m/s

Uncertainty in momentum is,

\Delta p=0.2\%\ of\ 24.81

\Delta p=0.049

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

\Delta p.\Delta x\geq \dfrac{h}{4\pi}

\Delta x=\dfrac{h}{4\pi \Delta p}

\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}

\Delta x=1.07\times 10^{-33}\ m

Hence, this is the required solution.

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.
Maksim231197 [3]

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0

\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}

\frac{0.877GMm}{R} =\frac{-GMm}{R+h}

h = .14 R

b )

\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0

\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

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5 0
4 years ago
A 69-kg student sitting on a hardwood floor does not slide until pulled by a 260-N horizontal force. find coefficient of frictio
salantis [7]

Answer:

μ = 0.385

Explanation:

Given that,

The mass of the student, m = 69 Kg

The horizontal force applied, F = 260 N

The normal force acting on the body, weight = mg = 69g  N

                                                                                    = 676.2 N

The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.

The formula for coefficient of friction,

                                    μ =  F / N

Substituting the values in the above equation,

                                     μ = 260 N / 676.2 N

                                        = 0.385

Hence, the  coefficient of friction, μ = 0.385

6 0
3 years ago
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