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Resolving forces

Authored by Erell Bonnot

Physics

10th Grade - University

NGSS covered

Used 69+ times

Resolving forces
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12 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

An aeroplane is flying horizontally and heading north through the air. Its speed through the air is a and the wind is blowing east with a speed b.


The speed over the ground is given by

a+ba+b

a2+b2a^2+b^2

a+b\sqrt{a+b}

a2+b2\sqrt{a^2+b^2}

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

With reference to that same aeroplane, the angle from North at which the plane flies over the ground is given by

cos1(ab)\cos^{-1}\left(\frac{a}{b}\right)

sin1(ba)\sin^{-1}\left(\frac{b}{a}\right)

tan1(ab)\tan^{-1}\left(\frac{a}{b}\right)

tan1(ba)\tan^{-1}\left(\frac{b}{a}\right)

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

The diagram shows the forces acting on a picture, of weight W, suspended by a cord.

The tension in the cord is T. Which of the following expressions shows the correct relationship between W and T?

W=2TcosθW=2T\cos\theta

W=TcosθW=T\cos\theta

W=TsinθW=T\sin\theta

W=2TsinθW=2T\sin\theta

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Read this one carefully:


The diagram shows the forces acting on an object on an inclined surface. The component of R parallel to the inclined surface is

0

W

RcosθR\cos\theta

RsinθR\sin\theta

Tags

NGSS.HS-PS2-1

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

With this one, try drawing the diagram yourself to help you if you need it:

A football is kicked at 12 m s-1 at an angle of 35° to the horizontal. The horizontal component of its velocity, in m s-1, is given by

12cos 35°12\cos\ 35\degree

12sin35°12\sin35\degree

12cos35°\frac{12}{\cos35\degree}

12sin35°\frac{12}{\sin35\degree}

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Again, try drawing the diagram yourself here to help you if you need to:

A model boat is crossing a stream. The stream is travelling east at a speed of 1.5 m s1s^{-1}  .The boat is heading north at a speed of 0.5 m s1s^{-1}  . The magnitude of the resultant velocity is given by

 (1.52+0.52) m s1\left(1.5^2+0.5^2\right)\ m\ s^{-1}  

 (1.52+0.52)  m s1\sqrt{\left(1.5^2+0.5^2\right)}\ \ m\ s^{-1}  

 (1.5 +0.5) m s1\left(1.5\ +0.5\right)\ m\ s^{-1}  

 (1.5+0.5) m s1\sqrt{\left(1.5+0.5\right)}\ m\ s^{-1}  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Remember, do a diagram if you think it will help:


A body is acted on by a vertical force of 18 N and a horizontal force of 32 N.


The angle to the horizontal of the resultant force is given by

cos1(1832)\cos^{-1}\left(\frac{18}{32}\right)

tan1(1832)\tan^{-1}\left(\frac{18}{32}\right)

sin1(3218)\sin^{-1}\left(\frac{32}{18}\right)

tan1(3218)\tan^{-1}\left(\frac{32}{18}\right)

Tags

NGSS.HS-PS2-1

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