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4 years ago

Select all that apply.

Physics

2 answers:

sashaice [31]4 years ago

8 0

Answer:

-left-right reversed

-upright

-virtual

I just did this in my lesson.

Pani-rosa [81]4 years ago

6 0

The correct options are:

- upright (the image of a flat mirror always appear in the same vertical-orientation of the object)
- left-right reversed (the image of a flat mirror always appear inverted in the horizontal direction)
- virtual (the image appears in fact behind the plane of the mirror)

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Driving in a straight line at 60 miles per hour

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3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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1 year ago

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IMA stands for ideal mechanical advantage, which is the theoretical force amplification factor on an ideal mechanical device free of friction, deformations, etc.

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3 years ago

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