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galben [10]
3 years ago
8

A student leaves a sidewalk corner walks 3.0 blocks north then walks 3.0 blocks west.what is the students displacement from the

starting position
Physics
1 answer:
Sliva [168]3 years ago
3 0

Answer:

The students displacement is 3·√3 blocks, 45° North of West

Explanation:

From the question, we have;

The distance North in which the student walks = 3.0 blocks

The next walking distance West by the student = 3.0 blocks

The displacement of the student = The shortest distance between the student start and final point

By representing the motion of the student in vector format, we have;

\mathbf{d} = 3\cdot \hat{i} + 3\cdot \hat{{j}}

Therefore the magnitude of d is given as follows;

\left | d\right | = \sqrt{3^2 + 3^2} = 3\cdot \sqrt{3} \ blocks

The direction of the student = tan⁻¹(3/3) = 45° West of North

The students displacement = 3·√3 blocks, 45° North of West.

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The act of using senses or tools to gather information is called <em>Obser</em><em>vation</em><em>.</em>

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The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th
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Answer:

Momentum, p=\hbar k

Explanation:

The wave function of a particle is given by :

y=exp[i(kx-\omega t)]...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

\omega is the angular frequency

The relation between the momentum and wavelength is given by :

p=\dfrac{h}{\lambda}............(2)

From equation (1),

k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{k}

Use above equation in equation (2) as :

p=\dfrac{h k}{2\pi }

Since, \dfrac{h}{2\pi}=\hbar

p=\hbar k

So, the x-component of the momentum of the particle is \hbar k. Hence, this is the required solution.

8 0
3 years ago
Consider a particle with unit charge q, and mass m, in a constant magnetic field B directed along the positive z–axis. The parti
max2010maxim [7]

Answer:

it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

Explanation:

Magnetic field is along z axis while velocity is in x-z plane

so we will have

F = q(\vec v \times \vec B)

so here we can say

F = q(u\hat i + w\hat k) \times (B \hat k)

so we will have

F = quB(-\hat j)

so here the net force on the charge is perpendicular to its x directional velocity along - Y direction

So due to this component of motion it will move along a circle while other component of the motion will remain uniform always

So here it is combination of two parts

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So we can say that it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

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someone help me please haha Calculate the braking force of a lorry decelerating at a rate of 8 m/s² with a mass of 5000 kg.
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Answer:

40000 N

Explanation:

Force: This cam be defined as the product of mass and acceleration.

From the question,

The braking force of the lorry will be the same in magnitude as the force at which the lorry is moving.

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Where m = mass of the lorry, a = acceleration/deceleration of the lorry.

Given: m = 5000 kg, a = 8 m/s²

Substitute these values into equation 1

F = 5000(8)

F = 40000 N

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I’m sure gravitation should be at the river sorry if I’m wrong
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