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KengaRu [80]
4 years ago
11

Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or

der to get the student-student-skateboard system to accelerate, Student X claims that Student Y should apply a force on Student X while both students stand on the skateboard. Which of the following statements is true regarding the claim made by Student X? a)The claim is correct because Newton’s second law states that an object will accelerate if a net force is applied to the object. b)The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system. c)The claim is incorrect because Student Y cannot apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system. d)The claim is incorrect because both students are internal to the student-student-skateboard system, and internal forces within a system cannot cause the system to accelerate.
Physics
2 answers:
OleMash [197]4 years ago
6 0

Answer:

the answer is B.

Explanation:

The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system

stellarik [79]4 years ago
5 0

Answer:

The claim is incorrect because both students are internal to the student-student-skateboard system, and internal forces within a system cannot cause the system to accelerate.

Explanation:

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Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
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Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

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Explanation:

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Displacement is the change in velocity of an object.<br><br> True or false
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Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate
Akimi4 [234]

Answer:

a) The position vector of P is \vec P =(0, 2,4).

b) The distance vector from P to Q is \overrightarrow{PQ} = (-3,-1,1).

c) The distance between P and Q is \|\overrightarrow{PQ}\|=\sqrt{11}.

d) A vector parallel to PQ with magnitude of 10 is \vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right).

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

\vec P = (0,2,4)-(0,0,0)

\vec P = (0-0, 2-0, 4-0)

\vec P =(0, 2,4)

The position vector of P is \vec P =(0, 2,4).

b) First, we calculate the position vector of point Q:

\vec Q = (-3,1,5)-(0,0,0)

\vec Q = (-3-0,1-0,5-0)

\vec Q =(-3,1,5)

The distance vector from P to Q is define by the following vectorial expression:

\overrightarrow{PQ} = \vec Q - \vec P (1)

\overrightarrow{PQ} = (-3,1,5)-(0,2,4)

\overrightarrow{PQ} =(-3-0,1-2,5-4)

\overrightarrow{PQ} = (-3,-1,1)

The distance vector from P to Q is \overrightarrow{PQ} = (-3,-1,1).

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}

\|\overrightarrow{PQ}\|=\sqrt{11}

Second Method - Dot Product:

\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}} (2)

\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}

\|\overrightarrow{PQ}\|=\sqrt{11}

The distance between P and Q is \|\overrightarrow{PQ}\|=\sqrt{11}.

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ} (3)

Where k is the scale factor.

If we know that \overrightarrow{PQ} = (-3,-1,1), \|\overrightarrow{PQ}\|=\sqrt{11} and k = 10, then the vector is:

\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)

\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)

\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)

A vector parallel to PQ with magnitude of 10 is \vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right).

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3 years ago
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