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3 years ago

A 9mm bullet has a mass of 0.007 kg. When fired

Physics

1 answer:

Sonbull [250]3 years ago

3 0

Answer:

The velocity of the bullet is 380 m/s

Explanation:

The given parameters are;

The mass of the 9 mm bullet fired = 0.007 kg

The measure of the momentum the bullet has = 2.66 kg·m/s

The equation for linear momentum is given as follows;

The linear momentum of an object = The mass of the object × The velocity of the object

∴ The velocity of the object = The linear momentum of an object/(The mass of the object)

The velocity of the bullet = The linear momentum of an bullet/(The mass of the bullet)

The velocity of the bullet = 2.66 kg·m/s/(0.007 kg) = 380 m/s

The velocity of the bullet = 380 m/s.

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Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron's velo

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Answer:

Velocity of electron along x direction is 57.9 m/s

Explanation:

The uncertainty in x coordinate of electron, Δx = 0.200 mm = 0.2 x 10⁻³ m

Let vₓ be the x component of electrons velocity.

The uncertainty in x component of electrons momentum is:

Δpₓ = mΔvₓ

Here m is mass of the electron.

The uncertainty in velocity x component is 1% i.e. 0.01.

So, the above equation can be written as :

Δpₓ = 0.01mvₓ ....(1)

The minimum uncertainty principle is:

$\Delta x\Delta p_{x} = \frac{h}{2\pi }$ ....(2)

Here h is Planck's constant.

From equation (1) and (2),

$\Delta x\times0.01m v_{x} = \frac{h}{2\pi }$

Substitute 0.2 x 10⁻³ m for Δx, 9.1 x 10⁻³¹ kg for m and 6.626 x 10⁻³⁴ m²kg/s in the above equation.

$0.2\times10^{-3} \times0.01\times9.1\times10^{-31}\times v_{x} = \frac{6.626\times10^{-34} }{2\pi }$

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3 years ago

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches

yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

$H = ut - \frac{1}{2}gt^{2}$

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴ $-63 = 0(t) - \frac{1}{2}(0.98) t^{2} \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s$

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

$v = u - gt$

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴ $v = - 9.8(3.59)\\v = - 35.18m/s$

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

$v = u + at$

Here, the initial velocity will be the final velocity just before the parachute opens, that is

$u = - 35.18 m/s$

From the question,

$v = - 3.3 m/s$

$a = 1.5 m/s^{2}$

We then get

$-3.3 = - 35.18 + (1.5)t$

$-3.3 + 35.18 = 1.5t\\31.88 = 1.5t$

$t = \frac{31.88}{1.5}$

$t = 21.25 secs$

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

$x = ut + \frac{1}{2}at^{2} \\$

$x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2} \\x = -747.58 + 338.67\\x = -408.91m\\$

$x$ ≅ $- 409m$

Hence, the height the fall begins is 63m + 409m

= 472 m

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Answer:

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