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Komok
3 months ago
6

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

f it takes the cart 1.5 seconds to reach 8.2 cm/s, what is the acceleration of the cart? Round your answer to the nearest tenth. cm/s2
Physics
2 answers:
serg [3.5K]3 months ago
5 0
First, we need to establish the motion equations for the cart.
This leads us to:
vf = a * t + vo
By substituting the values, we have:
8.2 = a * (1.5) + (3.5)
Solving for acceleration gives:
 a = (8.2-3.5) / (1.5)
 a = 3.1 m / s ^ 2
Response:
The acceleration of the cart is:
a = 3.1 m / s ^ 2
serg [3.5K]3 months ago
5 0

Answer:

3.1 cm/s²

Explanation:

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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
Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

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

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

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

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

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

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

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

According 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 widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
3 months ago
Is velocity ratio of a machine affected by applying oil on it?Explain with reason.​
Yuliya22 [3333]
Factors influencing friction

The magnitude of friction is contingent on the following elements: i) The surface area in contact. ii) The applied pressure on the surfaces. Force is determined by Pressure multiplied by Area; thus, if the contact area increases or if the pressure applied rises, the frictional force will also escalate.

Methods for reducing friction

i) Smooth the contact surface. ii) Apply oil or grease to fill small gaps in flat surfaces. iii) Use ball bearings to minimize contact area among rotating components.

Lubrication

To minimize friction, various methods may be employed: Oil can be either thin or viscous, which depends on its SAE number (SAE indicating Society of Automotive Engineers). Highly viscous oils may not reach all components effectively. In contrast, very thin oils may drain away quickly, resulting in wastage. Grease is preferable in such situations, particularly around ball-bearings. Regular grease or oil should not be utilized under high speed, high pressure, and high temperature conditions—specialized lubricants are required then. The consistency of oil varies with temperature; it thickens in the cold and thins in the heat. Therefore, the choice of lubricant should be seasonally appropriate, and it's always wise to consult the equipment's operating manual prior to making a selection.[[TAG_11]]
6 0
3 months ago
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