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jeyben [28]
3 years ago
13

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

nelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground.
What is the average x component of the acceleration of the car during the collision?
What is the average x component of the acceleration of the truck during the collision?
Physics
1 answer:
tresset_1 [31]3 years ago
5 0

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

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Physics B 2020 Unit 3 Test
weqwewe [10]

Answer:

1)

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F=qvB sin \theta

where here:

For the proton in this problem:

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So the force is

F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

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- The density of field lines is lower far from the Poles

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3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

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4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

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v=2155 m/s is the speed of the alpha particle

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r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

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The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

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2- A circular motion in the direction perpendicular to the magnetic field

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8)

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\Phi = BA sin \theta

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So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

Download docx
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