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san4es73 [151]
3 years ago
14

MATHS PROBLEM SOLVING

Mathematics
1 answer:
lukranit [14]3 years ago
3 0
Very nice Question! Please be careful during each step:

1) First fill 5 liter bucket, and transfer it's water to 3 liter. Fill 3-liter bucket completely from it, so now you have 2 liter water in 5 liter bucket.

2) Now, empty 3 liter bucket, & transfer that 2 liter water from 5 liter bucket to it.
Now, you have an empty 5 liter bucket + 2 liter water in 3 liter bucket.

3) Now, fill 5 liter bucket completely, and fill 3-liter bucket completely, which had 1 liter space empty, so, Now, you have 5 - 1 = 4 liter in 5 liter bucket.

You Got that!! Congrats!! If You 're thirsty by that hard work, then you can drink that water, if it is clean!! :)

Hope this helps!
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Toni had 5 1/2 kilograms of sugar.how many cakes can he bake if he ise 1/4 hilogram of sugar per cake?
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I need help with Q9 c, I’ve done a and b if it helps, b=325 but I am just stuck on question 9C, I know the answer is 500N (as it
777dan777 [17]

Step-by-step explanation:

I know you've already done parts a and b, but I'll show the work for that before I do c.

Draw two free body diagrams, one for the car and one for the trailer.  The car pulls the trailer forward with a tension force T, so the trailer pulls backward on the car with an equal and opposite force T.

The car also has a 1200 N forward force from the engine, and a 200 N backwards force from resistance.

The trailer has a backwards resistance force of 100 N.

Sum of forces on the car:

∑F = ma

1200 − 200 − T = 900a

1000 − T = 900a

Sum of forces on the trailer:

∑F = ma

T − 100 = 300a

To solve the system of equations, first add the equations together.

1000 − 100 = 1200a

900 = 1200a

a = 0.75 m/s²

Plug back into either equation to find the tension force:

T = 325 N

Now for part c, draw new free body diagrams for the car and trailer.  This time, the car is pushing back on the trailer to slow it down.  So the trailer is pushing forward on the car with an equal and opposite force.  The magnitude of that tension force is given to be 100 N.

The car also has a backwards 200 N force from resistance, and a backwards brake force F.

The trailer has a backwards 100 N force from resistance.

Sum of forces on the car:

∑F = ma

100 − 200 − F = 900a

-100 − F = 900a

Sum of forces on the trailer:

∑F = ma

-100 − 100 = 300a

-200 = 300a

a = -⅔

Plugging into the first equation:

-100 − F = 900 (-⅔)

-100 − F = -600

F = 500 N

4 0
2 years ago
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