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Alenkasestr [34]
3 years ago
13

QUESTION 1

Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years

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Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1). Graph △XYZ and its image after a rotation of 180° about (2, –3).
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The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).

If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

More about the geometry link is given below.

brainly.com/question/7558603

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Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

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