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Gnom [1K]
4 years ago
12

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:

<span>- upright (the image of a flat mirror always appear in the same vertical-orientation of the object)
</span><span>- left-right reversed (the image of a flat mirror always appear inverted in the horizontal direction)
</span><span>- virtual (the image appears in fact behind the plane of the mirror)</span>
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Which of the following represents a case in which you are not accelerating? -Driving 60 miles per hour around a curve -Going fro
PIT_PIT [208]

Answer:

Driving in a straight line at 60 miles per hour

Explanation:

In the first case there's an acceleration that modifies the direction of the movement.

In the second case there's a lineal acceleration that increases the speed of the car.

in the third case there's a negative acceleration that reduces the speed of the car.

On the third case the speed is constant so the acceleration is 0 mi/s^2

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3 years ago
What is the force when two charged spheres distance is in half​
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When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres

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4 years ago
A hair dryer required 12 V to operate properly. When plugged into a 120 V outlet, a transformer must change voltage. If the prim
grigory [225]

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5 0
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.

8 0
1 year ago
If the IMA of a machine is 2 and the effort force is 50 newtons, then the force applied to the resistance is _____. 48 N 50 N 25
zysi [14]

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


If the applied force (effort) is 50N, then the force applied to the resistance is multiplied by the IMA=2 to get 100N.

7 0
3 years ago
Read 2 more answers
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