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patriot [66]
4 years ago
9

Can someone check my answers and tell me if their correct?

Physics
1 answer:
Otrada [13]4 years ago
6 0

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read.  the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi =  2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 =  (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0  m/s   But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11


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Now you can see the effects of changing F1, F2, x1 and x2.

If you decrease the lengt X1 between the applied effort (F1) and the pivot,  IMA decreases.

If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.

If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.

If you decrease the applied effort (F1) and decrease the distance between it and the pivot  (X1) IMA will decrease.

Answer: Increase the length between the applied effort and the pivot.
4 0
4 years ago
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A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t
Readme [11.4K]

Answer:

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Explanation:

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Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

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3 years ago
Think of an example that you could use to convice a friend that an object at rest has an "internal hold back property" that is d
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Answer:

Explanation:

There are various examples around us that states " internal hold back property " and is different from frictional force or the force of Earth's gravity.

The plaster which is the mixture of cement and water has an adhesive nature and molecules inside them has a internal hold back property. They stick to the walls and ceiling and do not fall.

Though gravity is applied on it still its molecules tends to hold the ceiling firmly and do not fall upon.

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4 years ago
As a motorcycle takes a sharp turn, the type of motion that occurs is called _______________ motion.
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Answer:

circular motion

Explanation:

As a motorcycle takes a sharp turn, the type of motion that occurs is called circular motion.

Circular motion is a movement of an object along a circular path. As this motorcycle makes the sharp turn, it is acted upon by a centripetal force which directs the motorcycle towards the center.

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3 years ago
A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
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Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

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We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

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We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

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Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0
3 years ago
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