Abby5001
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This should be simple but I am really struggling. can you explain why the answer is 22m/s and 38 degrees??,Name:  image.jpg
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Discipulus
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Add the two vectors on the diagram by putting them end to end (making sure they're at the correct angles) and drawing a line from the start of the first vector to the end of the last vector, this is the total velocity.
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uberteknik
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(Original post by Abby5001)
This should be simple but I am really struggling. can you explain why the answer is 22m/s and 38 degrees??,Name:  image.jpg
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Not enough information provided.

You need to post the whole question for a definitive answer. It's pretty certain this is a vector addition problem, but without all of the information it's guesswork to try and answer you.
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Discipulus
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(Original post by uberteknik)
Not enough information provided.

You need to post the whole question for a definitive answer. It's pretty certain this is a vector addition problem, but without all of the information it's guesswork to try and answer you.
There is enough information; draw the first vector at 65 degrees and then the next one horizontal to the page (like in the diagram) joining on to it at the correct magnitudes.
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Abby5001
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(Original post by uberteknik)
Not enough information provided.

You need to post the whole question for a definitive answer. It's pretty certain this is a vector addition problem, but without all of the information it's guesswork to try and answer you.
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uberteknik
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(Original post by Abby5001)
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Thanks.

Ignoring the river velocity for a moment, if the new boat heads off at 1.5 m/s at an angle of 65o to the riverbank, then resolving the horizontal and vertical components produces:

\mathrm{horizontal \ velocity} = 1.5cos(65) = 0.634 \mathrm{\ m/s}

\mathrm{vertical \ velocity} = 1.5sin(65) = 1.36 \mathrm{\ m/s}

This is in addition to the river velocity of 1.1 m/s

i.e. the boat must actually travel at

1.36 m/s across the river and,

1.1 + 0.634 = 1.734 m/s downstream

Combining these produces the resultant velocity of:

\sqrt{(1.36^2 + 1.734^2)} = 2.2 \mathrm{\ m/s}

at an angle of:

\theta = tan^{-1}\frac{1.36}{1.734} = 38^o
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Abby5001
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(Original post by uberteknik)
Thanks.

Ignoring the river velocity for a moment, if the new boat heads off at 1.5 m/s at an angle of 65o to the riverbank, then resolving the horizontal and vertical components produces:

\mathrm{horizontal \ velocity} = 1.5cos(65) = 0.634 \mathrm{\ m/s}

\mathrm{vertical \ velocity} = 1.5sin(65) = 1.36 \mathrm{\ m/s}

This is in addition to the river velocity of 1.1 m/s

i.e. the boat must actually travel at

1.36 m/s across the river and,

1.1 + 0.634 = 1.734 m/s downstream

Combining these produces the resultant velocity of:

\sqrt{(1.36^2 + 1.734^2)} = 2.2 \mathrm{\ m/s}

at an angle of:

\theta = tan^{-1}\frac{1.36}{1.734} = 38^o
thankyou so much!!! I really appreciate it😀😀
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uberteknik
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(Original post by Discipulus)
There is enough information; draw the first vector at 65 degrees and then the next one horizontal to the page (like in the diagram) joining on to it at the correct magnitudes.
Yes of course, it was me misreading the question!
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Abby5001
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Another question I am struggling with. why would you do cos65 when the angle the 25 degrees to the horizontal
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uberteknik
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(Original post by Abby5001)
Another question I am struggling with. why would you do cos65 when the angle the 25 degrees to the horizontal
We need to resolve the forces acting on the geologist (in a similar way to resolving the velocity components of the last example) using a reference that puts the forces at right angles to each other and in the direction the geologist must travel up the slope.





Since the geologist is on a mountain slope and not on horizontal ground, he can only move up the slope (hypotenuse), so the forces of gravity and friction need to be resolved. There are two resolved component forces acting at right angles: - the normal force perpendicular to the slope and the frictional force parallel to the slope.

Using the trigonometry rules, we can draw the graphical representation of the force magnitudes and directions:




At this point, we need to introduce the trigonometry rules for the angles of similar triangles. Notice where theta occurs (three places but only two shown explicitly in the diagram above).

Notice that the actual force of gravity acts vertically downwards so the magnitude of that force is given by:

F_g = mg

Which makes the normal (to the slope) perpendicular component force:

F_n = F_gCos\theta

And the frictional parallel (to the slope) component force:

F_f = F_gSin\theta
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Abby5001
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(Original post by uberteknik)
We need to resolve the forces acting on the geologist (in a similar way to resolving the velocity components of the last example) using a reference that puts the forces at right angles to each other and in the direction the geologist must travel up the slope.





Since the geologist is on a mountain slope and not on horizontal ground, he can only move up the slope (hypotenuse), so the forces of gravity and friction need to be resolved. There are two resolved component forces acting at right angles: - the normal force perpendicular to the slope and the frictional force parallel to the slope.

Using the trigonometry rules, we can draw the graphical representation of the force magnitudes and directions:




At this point, we need to introduce the trigonometry rules for the angles of similar triangles. Notice where theta occurs (three places but only two shown explicitly in the diagram above).

Notice that the actual force of gravity acts vertically downwards so the magnitude of that force is given by:

F_g = mg

Which makes the normal (to the slope) perpendicular component force:

F_n = F_gCos\theta

And the frictional parallel (to the slope) component force:

F_f = F_gSin\theta
If I'm being honest I am still confused.
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uberteknik
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(Original post by Abby5001)
If I'm being honest I am still confused.
Hmmmm. Diagram seems to be copyright protected.

Take a look at this video which explains the concept:

https://youtu.be/dA4BvYdw7Xg
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Abby5001
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Doing some physics past papers and I need help:Can someone tell me why the answers respectively are:

2R
R/2
2R/3
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lmsavk
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(Original post by Abby5001)
Doing some physics past papers and I need help:Can someone tell me why the answers respectively are:

2R
R/2
2R/3
A open, B closed: No current flows through the A branch, because switch A is not closed... it's a broken path. B is closed so current will flow through the first resistor in branch B, but not the second, because it will take the path of least resistance through the parallel connection with the second resistor in B branch.

A open, B open: The current cannot flow through A branch, as above. The current cannot flow through the parallel connection across the second resistor of B branch. Therefore, it must flow through both resistors on B branch, they're in series so we add the resistances together: 2R.

A closed, B closed: With both switches closed, the current will flow through the A branch and also the first resistor of the B branch. It won't flow through the second resistor of the B branch because switch B is closed, the current will take the path of least resistance. The two resistors form a parallel connection with one another:
\frac{1}{R_{total}}=\frac{1}{R_1  }+\frac{1}{R_2}

Knowing the general format of the answers, you should be able to deduce the last. It's a simple case of knowing where the current is flowing - that should always be on your mind when you look at a circuit.
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Sacho12
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That is the whole question
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