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denis-greek [22]
3 years ago
7

Help on these two thanks

Physics
2 answers:
egoroff_w [7]3 years ago
6 0

Answer:

8. Friction

9. Net force

Explanation:

Hope this helps you. Have a nice day. :)

OLEGan [10]3 years ago
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Serga [27]
Negative Reinforcement
5 0
3 years ago
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
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 65 kg and exerts
Leviafan [203]

Answer:

.team 1 a = 20.8 m/s²   team 2  a = 19.2 m/s²

Explanation:

Let's use Newton's second law to calculate the acceleration of the two groups

   

      F = ma

      a = F / M

Group 1

      a = 9 1354/9 65

      a = 20.8 m / s²

Group 2

     a = 9 1364/9 71

     a = 19.2 m / s²

If the two groups pull against each other, group 1 should win, by creating a greater acceleration.

     Δa = 20.8 -19.2

     Δa = 1.6 m / s²

4 0
4 years ago
Which type of bond is found between the atoms of a molecule?
Brrunno [24]

Answer:

Covalent Bond is found between the atoms of a molecule.

4 0
3 years ago
Question #4
olga2289 [7]

Answer:

Distance 5 km, Displacement 3 km east

Explanation:

6 0
3 years ago
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