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4 years ago

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A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The b

ullet embeds into the block. The speed, in m/s, of the bullet-plus-wood combination immediately after the collision is

1 answer:

ch4aika [34]4 years ago

6 0

Answer:

1.77 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............................ Equation 1

Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

make V the subject of the equation

V = (mu+m'u')/(m+m').................. Equation 2

Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)

Substitute into equation 2

V = (0.04×310+6.96×0)/(0.04+6.96)

V = 12.4/7

V = 1.77 m/s

Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

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A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

4 years ago

What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL

pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of object = $6.0g/ml$

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

$6.0g/ml=\frac{3.0g}{\text{Volume of substance}}$

${\text{Volume of substance}}=0.50ml$

Thus the volume of an irregularly shaped object is 0.50 ml

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I believe it is call “Acceleration”

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zzz [600]

Answer:

●1m to KM

Divided by 1000

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