# Linearising equations question

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

I'm slightly stuck on these two questions, anyone got any hints?

Last edited by Slx.24; 4 weeks ago

0

reply

Report

#2

wouldn't the first one have to be rearranged so that the s is the denominator and Ek the numerator and then substitute -0.98?

0

reply

Report

#3

(Original post by

I'm slightly stuck on these two questions, anyone got any hints?

**Slx.24**)I'm slightly stuck on these two questions, anyone got any hints?

In this case substitute the value for W in: Ek = -2.39 sin(θ) s + E.

Then set -2.39 sin(θ) = -0.98 as we can see that -Wsin(θ) is the gradient and we are given the value of the gradient. Remember it is -2.39 not 2.39 as it is -W.

This equation can be very simply solved (I hope!): -2.39 sin(θ) = -0.98 (solve for θ)

2) Again remember the equation of a straight line and we can see that the gradient which in this case is (e/h) = 2.41x10^14 and we know the value of e. It should be quite easy to find the value of h.

And the same can be done for the intercept to find the other value.

0

reply

(Original post by

1) I recommend remembering the equation of a straight line which is y = mx+ c, where m is the gradient and c is the y intercept. It may also help if you see it as Ek = -Wsin(θ) s + E0.

In this case substitute the value for W in: Ek = -2.39 sin(θ) s + E.

Then set -2.39 sin(θ) = -0.98 as we can see that -Wsin(θ) is the gradient and we are given the value of the gradient. Remember it is -2.39 not 2.39 as it is -W.

This equation can be very simply solved (I hope!): -2.39 sin(θ) = -0.98 (solve for θ)

2) Again remember the equation of a straight line and we can see that the gradient which in this case is (e/h) = 2.41x10^14 and we know the value of e. It should be quite easy to find the value of h.

And the same can be done for the intercept to find the other value.

**Zain-**)1) I recommend remembering the equation of a straight line which is y = mx+ c, where m is the gradient and c is the y intercept. It may also help if you see it as Ek = -Wsin(θ) s + E0.

In this case substitute the value for W in: Ek = -2.39 sin(θ) s + E.

Then set -2.39 sin(θ) = -0.98 as we can see that -Wsin(θ) is the gradient and we are given the value of the gradient. Remember it is -2.39 not 2.39 as it is -W.

This equation can be very simply solved (I hope!): -2.39 sin(θ) = -0.98 (solve for θ)

2) Again remember the equation of a straight line and we can see that the gradient which in this case is (e/h) = 2.41x10^14 and we know the value of e. It should be quite easy to find the value of h.

And the same can be done for the intercept to find the other value.

Thanks! I did like 90% of that but I got stuck because I thought there were two unknowns, one being theta (sorry on a phone rn) I knew what the gradient was and did equate the whole thing to -0.98 but just couldn't figure the whole thing out because I thought 0.98= E0 -2.39 sin(theta) BUT we don't know what E0 is as well as sin(theta) so I was completely stumped.

Are you saying that E0 is somehow not part of the gradient?

(Original post by

wouldn't the first one have to be rearranged so that the s is the denominator and Ek the numerator and then substitute -0.98?

**englishhopeful98**)wouldn't the first one have to be rearranged so that the s is the denominator and Ek the numerator and then substitute -0.98?

0

reply

Report

#5

(Original post by

(Original post by Zain-)

Thanks! I did like 90% of that but I got stuck because I thought there were two unknowns, one being theta (sorry on a phone rn) I knew what the gradient was and did equate the whole thing to -0.98 but just couldn't figure the whole thing out because I thought 0.98= E0 -2.39 sin(theta) BUT we don't know what E0 is as well as sin(theta) so I was completely stumped.

Are you saying that E0 is somehow not part of the gradient?

Idts, the x and y values were in the right order so you dont need to rearrange further (for the first one anyways) s = x and Ek = y which means the things in the middle is the gradient

**Slx.24**)(Original post by Zain-)

Thanks! I did like 90% of that but I got stuck because I thought there were two unknowns, one being theta (sorry on a phone rn) I knew what the gradient was and did equate the whole thing to -0.98 but just couldn't figure the whole thing out because I thought 0.98= E0 -2.39 sin(theta) BUT we don't know what E0 is as well as sin(theta) so I was completely stumped.

Are you saying that E0 is somehow not part of the gradient?

Idts, the x and y values were in the right order so you dont need to rearrange further (for the first one anyways) s = x and Ek = y which means the things in the middle is the gradient

Ek = -Wsin(θ) s + E0.

This is in the form y = mx + c where m is the gradient, c is the y intercept, y and x are both axes.

Also to note, the gradient is always multiplying the x axis value (in this case s), as seen from the general equation above. We can see very clearly that E0 is not multiplying s, so therefore it is not part of the gradient. However, if the question had brackets like this: Ek = (E0 - Wsinθ) s, then here is an example of where E0 is part of the gradient but the question does not ask this.

Also, it is important to understand that the gradient is not "the things in the middle". I have explained what the gradient is directly above if you need to look back. The equation can be given in any order so it is important to understand what each part of the equation is as you will be required to rearrange it into the correct order - generalised by y = mx + c.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top