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Romashka [77]
4 years ago
13

A solid ball has a radius of 2cm and a length of 7cm. It has a density of 3.1g/cm3

Physics
1 answer:
Iteru [2.4K]4 years ago
4 0
Density is mass divided by volume, you would have to solve for the volume of the ball and rearrange the equation to density divided by volume equals mass
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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
3 years ago
How does the resulting cell at the end of asexual reproduction compare to the original cell
lina2011 [118]

1. At the end of asexual reproduction the resulting cell is an exact genetic copy of the original cell.

2. During inter phase the cell copies its DNA, and grows to an appropriate size where it is safe for the cell to split.

3. After meiosis, the resulting daughter cells are 4 genetically different haploid daughter cells.

4. During sexual reproduction, two haploid gametes join in the process of fertilization to produce a diploid zygote. The two haploid gamete bring different traits together, they also shares some DNA. This makes each daughter cell unique.

4 0
3 years ago
would the phases of the moon be affected if the moon didn't make one rotation for each revolution of earth?
Aleks [24]

As long as the moon orbits the Earth in the same way, the phases will be the same.

4 0
3 years ago
In figure 16-2 is the temperature of the material within the cylinder greatest during the intake stroke, compression stroke, pow
erastovalidia [21]
Technically, we have no way of knowing that without seeing Figure 16-2.
So the question should be reported for incomplete content.  But I'm
going to take a wild stab at it anyway.

There's so much discussion of 'cylinder' and 'strokes' in the question,
I have a hunch that it's talking about the guts of a 4-stroke internal
combustion gasoline engine. 

If I'm right, then the temperature of the material within the cylinder is
greatest right after the spark ignites it.  At that instant, the material burns, 
explodes, expands violently, and drives the piston down with its stiff shot
of pressure.

This is obviously happening because of the great, sudden increase in
temperature when the material ignites and explodes.

It hits the piston with pressure, which leads directly to the power stroke.  
 
5 0
3 years ago
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon
jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

            Στ = 0

            W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

           x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

            x₂ = \frac{M_1g  }{m_2 g} \ x_1

           

let's calculate

           x₂ = \frac{100}{75}  \ 1

           x₂ = 1.33 m

7 0
3 years ago
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